NUMBERS:
RATIONAL
AND IRRATIONAL
IVAN NIVEN
2+V-5
3-4i
The L. W. Singer Company
New Mathematical Library
NUMBERS:
RATIONAL
AND
IRRATIONAL
COMPLEX NUMBERS
REAL NUMBERS
(REAL) ALGEBRAIC NUMBERS
NUMBERS CONSTRUCTIBLE BY
STRAIGHTEDGE AND COMPASS
RATIONAL NUMBERS
INTEGERS
I
NATURAL
NUMBERS
I
NEW MATHEMATICAL LIBRARY
'Published by
Random House and The L. W. Singer Company
for
the
Monograph Project
of
the
SCHOOL MATHEMATICS STUDY GROUPt
EDITORIAL
Mark Kac, Chairman (1961-1964)
Rockefeller Institute
PANEL
Anneli Lax, Technical Editor
New York University
E. G. Begle, Stanford University
L. Bers (1958-62; Chairman, 1958-1961), New York University
W. G. Chinn (1961-64), San Francisco Public &lwols
H. S. M. Coxeter (1958-61), University of Toronto
P. J. Davis (1961-64), National Bureau of Standards
P. R. Halmos (1958-63), University of Michigan
J. H. Hlavaty (1958-63), DeWitt Clinton High Sclwol
N. Jacobson (1958--61), Yale University
R. S. Pieters (1958-62), Phillips Academy
H. O. Pollak (1958-61), Bell Telephone Laboratories, Ir.c.
G. P6lya (1958-61), Stanford University
H. E. Robbins (1958-61), Columbia University
W. W. Sawyer (1958-60), Wesleyan University
N. E. Steenrod (1958-62), Princeton University
J. J. Stoker (1958-61), New York University
L. Zippin (1958-61), Queens College
t The
School Mathematics Study Group represents all parts of the mathe­
matical profession and all parts of the country. Its activities are aimed at the
improvement of teaching of mathematics in our schools. Further information
can be obtained from: School Mathematics Study Group, School
of Education, Stanford University, Stanford, California.
NUMBERS:
RATIONAL AND IRRATIONAL
by
Ivan Niven
University of Oregon
THE L. W .
1
SINGER COMPANY
A Division of Random House
Second Printing
©
Copyright,
196 1 , by Yale U niversity
A ll rights reserved under I nternational and P an-American Copyright
Conventions. Published in N ew York by Random House, I nc., and
simultaneously in Toronto, Canada, by Random House of Canada, L imited.
School Edition published by The L. W. Singer Company
Library of Congress Catalog Card N umber:
Manufactured in the United States of
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Note to the Reader
is one of a series written by professional mathematicians
Thisorderbookto make
some important mathematical ideas interesting and
in
understandable to a large audience of high school students and laymen.
Most of the volumes in the New Mathematical Library cover topics
not usually included i n the high school curriculum ; they vary in diffi­
culty, and, even within a single book, some parts require a greater degree
of concentration than others. Thus, while the reader needs little
technical knowledge to understand most of these books, he will have
to make an intellectual effort.
If the reader has so far encountered mathematics only i n classroom
work, he should keep in mind that a book on mathematics cannot be
read quickly. Nor must he expect to understand all parts of the book
on first reading. He should feel free to skip complicated parts and
return to them later ; often an argument will be clarified by a subsequent
remark. On the other hand, sections containing thoroughly familiar
material may be read very quickly.
The best way to learn mathematics is to do mathematics, and each
book includes problems, some of which may require considerable
thought. The reader is urged to acquire the habit of reading with paper
and pencil in hand ; i n this way mathematics will become increasingly
meaningful to him.
For the authors and editors this is a new venture. They wish to
acknowledge the generous help given them by the many high school
teachers and students who assisted in the preparation of these mono­
graphs. The editors are interested in reactions to the books in this
series and hope that readers will write to: Editorial Committee of the
NML series, in care of THE INSTITUTE OF MATHEMATICAL SCIENCES,
NEW YORK UNIVERSITY, New York 3, N.Y.
The Editors
NEW MATHEMATICAL LIBRARY
Other titles will be announced
as
ready
1. NUMBERS: RATIONAL AND IRRATIONAL by Ivan Niven
2. WHAT IS CALCULUS ABOUT? by W. W. Sawyer
3. INTRODUCTION TO INEQUALITIES by E. Beckenbach
and R. Bellman
4. GEOMETRIC INEQUALITIES by N. D. Kazarinoff
5. THE CONTEST PROBLEM BOOK, Problems from the Annual
High School Contests of the Mathematical Association of America,
compiled and with solutions by Charles T. Salkind
6. THE LORE OF LARGE NUMBERS by P. J. Davis
7. USES OF INFINITY by Leo Zippin
8. GEOMETRIC TRANSFORMATIONS by I. M. Yaglom, trans­
lated from the Russian by Allen Shields
vi
C O NT ENT S
3
Introduction
Chapter 1
1.1
1.2
1.3
1.4
1.5
1.6
Chapter 2
2.1
2.2
2.3
2.4
2.5
2.6
Chapter 3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Natural Numbers and Integers
9
10
11
13
15
18
19
Rational Numbers
21
21
23
26
30
34
36
Primes
Unique Factorization
Integers
Even and Odd Integers
Closure Properties
A Remark on the Nature of Proof
Definition of Rational Numbers
Terminating andNon-terminating Decimals
The Many Ways of Stating and Proving Propositions
Periodic Decimals
Terminating Decimals Written as Periodic Decimals
A Summary
Real Numbers
The Geometric Viewpoint
Decimal Representations
The Irrationality of v'2
The Irrationality of V3
The Irrationality of V6 and v'2 + V3
The Words We Use
An Application to Geometry
A Summary
38
38
39
42
43
44
45
46
51
CONTENTS
viii
Chapter 4
4.1
4.2
4.3
4.4
4.5
Irrational
Numbers
Closure Properties
Polynomial Equations
Rational Roots of Polynomial Equations
Further Examples
A Summary
52
52
54
57
62
64
Chapter 5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
Trigonometric and Logarithmic
Numbers
Irrational Values of Trigonometric Functions
A Chain Device
Irrational Values of Common Logarithms
Transcendental Numbers
Three Famous Construction Problems
Further Analysis of �
A Summary
65
65
68
69
71
73
78
79
Chapter 6
6.1
6.2
6.3
6.4
The Approximation of Irrationals by Rationals
Inequalities
Approximation by Integers
Approximation by Rationals
Better Approximations
81
81
84
86
89
6.5
Approximations to within
94
6.6
6.7
Limitations on Approximations
A Summary
Chapter 7
7.1
7.2
7.3
7.4
7.5
7.6
The Existence of Transcendental
:2
99
102
Some Algebraic Preliminaries
An Approximation to ex
The Plan of the Proof
Properties of Polynomials
The Transcendence of ex
A Summary
103
104
107
108
109
111
113
Appendix A Proof That There Are Infinitely Many Prime
Numbers
115
Appendix B Proof of the Fundamental Theorem of Arithmetic
117
Appendix C Cantor's Proof of the Existence of Transcendental
Numbers
122
Answers and Suggestions to Selected Problems
129
Index
135
Numbers
NUMBERS:
RATIONAL
AND
IRRATIONAL
INT RO DU CT ION
The simplest numbers are the positive whole numbers,1,2,3,and so
on, used for counting.These are called natural numbers and have been
with us for so many millennia that the famous mathematician Kronecker
reputedly said: "God created the natural numbers; all the rest is the
work of man. "
The basic necessities of everyday life led to the introduction of com­
mon fractions like 1/2,2/ 3,5/ 4,etc.t Such numbers are called rational
numbers, not because they are "reasonable," but because they are ratios
of whole numbers.
We may think of the natural numbers as represented by dots along
a straight line (Fig. 1), each dot separated by one unit of length from
3
2
Figure 1
4
the previous one, as,for example, the number of inches along a tape
measure. We may represent rational numbers along the same straight
line (Fig. 2) and think of them as measuring fractions of length.
.l.�
2 3
i
i
5
I
1
4"
I
2
Figure 2
Much later, the Hindus invented the all-important number 0, and in
the beginning of modern times Italian algebraists invented negative
numbers.These may also be represented on a straight line, as shown
in Fig. 3.
t For reasons of typography, the fractions t, t, i, and others in this book often
appear with slanted bars, i.e. as 1/2, 2/3 , 5/4.
3
4
N U MBERS : RATIONA L A N D I RRATIONA L
-2
2
_'1
3"
o
Figure 3
,
,
2
When mathematicians talk about rational numbers, they mean posi­
tive and negative whole numbers (which can be represented as ratios,
e.g., 2 = 2/ 1 = 6/ 3, etc.), zero, and common fractions.The positive
and negative whole numbers and zero are also called integers, therefore
the class of rational numbers contains the class of integers.
The discovery that common fractions are not sufficient for the pur­
poses of geometry was made by the Greeks more than 2500 years ago.
They noticed, to their surprise and dismay, that the length of the
1
{Zl
Figure 4
diagonal of a square whose sides are one unit long (Fig. 4) cannot be
expressed by any rational number. (We shall prove this in Chapter
3.)Today we express this fact by saying that the square root of 2 (which,
according to the PythagoreanTheorem, is the length of the diagonal
of such a square) is an irrational number. What this means geometrically
is that there is no common unit of length, no common mesh however
fine, that can be put on both the side and the diagonal of a square a
whole number of times. In other words, there is no unit of length, no
matter how small, such that the side and the diagonal of a square are
multiples of that unit. For the Greeks, this was an awkward discovery,
because in many of their geometric proofs they had presumed that
given any two line segments, there would be a common unit of length.
Thus there was a gap in the logical structure of Euclidean geometry-an
incompleteness in the discussion of ratios and proportions of lengths.
In Section 3.7, we show how this gap may be closed and the theory of
proportion made complete.
Similarly, the circumference of a circle is an irrational mUltiple,
namely 7r, of the diameter.Other irrational numbers appear when we
try to evaluate some of the basic functions in mathematics. For example,
if we try to find the values of a trigonometric function,say sin x, when
x has the value 60°, we are led to the irrational number v3/ 2; similarly,
if we evaluate the logarithmic function log x, even for rational values
I N T R O D U C TIO N
5
of x, we usually are led to irrational numbers. Although the numbers
listed in tables of logarithmic and trigonometric functions are ostensibly
rational, actually they are only rational approximations of the true
values,which are irrational with few exceptions. Clearly, then, irrational
numbers occur in various natural ways in elementary mathematics.
The real numbers consist of all rational and irrational numbers, and
form the central number system of mathematics. In geometry, any
discussion of lengths, areas, or volumes leads at once to the real
numbers. Geometry affords, in fact, a simple intuitive device for de­
scribing the real numbers, i.e., the numbers required to measure all
possible lengths in terms of a given unit length. If again we consider the
representation of numbers as points along a straight line, we find that,
although any segment, no matter how small, contains infinitely many
rational points, there are many other points (such as v2, 71", etc.)
which measure lengths that cannot be expressed by rational numbers.
But once all real numbers are taken into account, every point on the
line corresponds to exactly one real number and every real number
corresponds to exactly one point on the line. The fact that aI/lengths
can be expressed as real numbers is known as the completeness property
of these numbers,and on this property depends the entire development
of mathematical analysis.
Thus the real numbers are of two kinds,the rational and the irrational.
There is another,much more recent separation of the real numbers into
two categories, the algebraic numbers and the transcendental numbers.
A real number is said to be algebraic if it satisfies some algebraic equa­
tion with integer coefficients. For example, v2 is an algebraic number
because it satisfies x2
2 = O. If a number is not algebraic, it is said
to be transcendental. From this definition,it is not clear that there are
any transcendental, i.e., non-algebraic, numbers. In 1851, the French
mathematician,Liouville, established that transcendental numbers exist.
Liouville did this by exhibiting certain numbers which he proved to
be non-algebraic. In Chapter 7 we shall follow Liouville's method to
establish the existence of transcendentals.
Later in the 19th century, it was proved that 71" is a transcendental
number, and this result settled an ancient geometric construction
problem known as "squaring the circle." This is discussed in Chapter
5. Another advance in the 19th century was made by Cantor, a German
mathematician, who established the existence of transcendental num­
bers by an entirely different approach. Although Cantor's method, in
contrast to Liouville's, does not exhibit a transcendental number in
explicit form, it has the advantage of demonstrating that, in a certain
sense, many more numbers are transcendental than algebraic. Such a
statement requires the comparison of infinite classes, since there are
-
6
NUM BERS:
RATlONAL
AND
IRRATlONAL
infinitely many algebraic numbers and infinitely many transcendental
numbers. These ideas are somewhat removed from the main discussions
in this book, so Cantor's proof of the existence of transcendental num­
bers is given in Appendix C.
The plan of the book is to present the natural numbers, integers,
rational numbers, and real numbers in the first three chapters. Then
in Chapter 4, a standard method is given for identifying irrational
numbers. Chapter 5 deals with the so-called trigonometric and loga­
rithmic numbers, that is, those numbers whose values are given approxi­
mately in tables of trigonometric and logarithmic functions. Chapter 6
treats the q uestion of how closely it is possible to approximate irrational
n umbers by the use of rational numbers. This chapter is more difficult
and more specialized than the earlier chapters. It is included to give
some readers an opportunity to explore mathematical arguments of a
new kind.
Chapter 7 and Appendix C offer two entirely independent proofs of
the existence of transcendental numbers, Chapter 7 by the method of
Liouville, Appendix C by the method of Cantor. The techniques are
markedly different and the reader will be well rewarded if he follows
each. The proof in Chapter 7 is laden with unavoidable technical de­
tails ; and, even more than in the earlier chapters, the reader will have
to use pencil and paper to follow the arguments. In fact, it is possible
that the reader may find Chapters I through 5 not too troublesome,
Chapter 6 rather difficult, and Chapter 7 virtually impossible. In such
a case, it is suggested that the reader postpone the study of Chapter 7
until he has more mathematical experience. On the other hand, any
reader who finds very little trouble in going through Chapters I to 5
might prefer to read Chapter 7 before Chapter 6. In fact, Chapter 7 is
independent of the rest of the book except for one well-known result
on inequalities given in Sect. 6. 1 .
Appendix C can be read independently of Chapter 7 except that the
factor theorem, Theorem 7.2, is needed. If the reader is not familiar
with the theory of sets, he will find the ideas of Appendix C very novel.
Appendix A, on the infinitude of prime numbers, is not essential to
the arguments developed in this book ; it is included because of its close
relation to the main topic and because this elegant proposition dates
back to Euclid. Appendix B on the fundamental theorem of arithmetic,
on the other hand, is essential to our arguments, particularly to those
of Chapters 4 and 5 ; the proof of this theorem has been relegated to an
appendix because it is somewhat lengthy and difficult in comparison
with the proofs in the first five chapters. The mathematically inexperi­
enced reader can accept the fundamental theorem of arithmetic on faith.
There are many exercises at the ends of sections ; the reader should
I N T R O D U CT I O N
7
try a good number of these to check his understanding of the text.
(Mathematics cannot be learned by watching the other fellow do it!)
Some of the problems are starred to indicate greater difficulty. The
reader should not necessarily be unhappy if he cannot solve all of these.
Often, his success depends on his mathematical maturity, that is, his
acquaintance with a fairly broad collection of mathematical procedures
from his other studies of mathematics. Answers to problems are given
at the end of the book and also some suggestions for solving a few of
the more difficult problems
The system of real numbers, rational and irrational, can be ap­
proached at any one of several levels of rigor. (The word "rigor" is used
technically in mathematics to denote the degree to which a topic is
developed from a careful logical standpoint, as contrasted with a more
intuitive position wherein assertions are accepted as correct because
they appear somewhat reasonable or self-evident.) Our purpose is to
present a first look at the subject, along fairly intuitive lines. Thus we
offer no axioms or postulates as a basis for the study. The prospective
mathematician into whose hands this book may find its way will one
day want to examine a careful axiomatic development of the real
number system. Why so? The reason is that our viewpoint here is so
descriptive that it leaves some basic questions unanswered. For example,
in Chapter 3 we say that the real numbers can be described in this way,
that way, and the other way. But how can we be certain that these
various ways are descriptions of the same system? To give a more
concrete example of a q uestion we do not answer in this book: How
do we know that v2". v3 = v6 or that Y;S. Y;7 = Y;35? To
answer such questions, a precise definition of operations for irrational
numbers must be given. This will not be done here since it is not so easy
as it might appear, and it is best to postpone this type of treatment until
the student not only has more mathematical skill but also has a greater
appreciation of the nature and meaning of mathematical proof. As the
American mathematician E. H. Moore said, "Sufficient unto the day
is the rigor thereof."
"The nature and meaning of mathematical proof!" It is not possible
here and now to give a precise description of what constitutes a proof,
and herein lies one of the most puzzling bugbears for the beginning
student of mathematics. If the nature of proof cannot be described or
formulated in detail, how can anyone learn it ? It is learned, to use an
oversimplified analogy, in the same manner as a child learns to identify
colors, namely, by observing someone else identify green things, blue
things, etc., and then by imitating what he has observed. There may be
failures at first caused by an inadequate understanding of the categories
or patterns, but eventually the learner gets the knack. And so it is with
8
N U MBERS : RATIO N A L A N D I R RATIONA L
the riddle of mathematical proof. Some of our discussions are intended
to shed light on the patterns of proof techniques,and so to acquaint the
reader with notions and methods of proof. Thus while we cannot give
any sure-fire recipe for what is and what is not a valid proof,we do say
some things about the matter, and hope that the reader, before he
reaches the end of this book, will not only recognize valid proofs but
will enjoy constructing some himself.
C H A PT E R O N E
Natural Numbers and Integers
The number system of mathematics begins with the ordinary numbers
used in counting,
1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12, . . . .
These are the positive whole numbers which are called the natural
numbers. The smallest natural number is 1 , but there is no largest
natural number, because regardless of how large a number is chosen,
there exist larger ones. Thus we say that there are infinitely many
natural numbers.
If any two natural numbers are added, the result will be a natural
number; for example, 4 + 4 = 8 and 4 + 7 = 1 1 . Similarly, if any
two are multiplied, the product will be a natural n um ber ; for example,
4 X 7 = 28. These two properties can be stated briefly by saying that
the natural numbers are closed under addition and closed under multi­
plication . In other words, if we have a collection of objects (say the set
of all natural numbers) and an operation (say addition) such that, no
matter on which members of our set we operate (say 4 and 7), the
result is again a member of the original collection, then we say that the
set is closed under that operation. Suppose we consider only the
numbers 1, 2, 3. This set of only three numbers is not closed under
addition because 1 + 3 = 4, and 4 is not a member of this set. When
we speak of the set of natural numbers we shall mean the set of all
natural numbers. If we wish to consider only some of them, we shall
specify which ones we include in our set. Thus we have seen that the
set of natural numbers is closed under addition but that the special set
consisting of only the three natural numbers 1,2, 3 is not.
The natural numbers are not closed under subtraction. In order to
see this, we need only show that not every subtraction of one natural
9
10
N U M BE R S : RATION A L A N D I R RATION A L
number from another yields a natural number. For example, if 7 is
subtracted from 4 the result, - 3, is not a natural number. Of course
if 4 is subtracted from 7, the result is the natural number 3 ; according
to our definition, however, we cannot say that a set of numbers is closed
under subtraction unless the result of every possible subtraction is
contained within that set. Similarly the natural numbers are not closed
under division, because if 4 is divided by 7 for example, the result is
the fraction 4/7 which is not a natural number.
It happens in many cases that two natural numbers can be divided
to give a natural number as a result, for example 35 divided by 5 gives
7. In this case we say that 5 is an exact divisor of 35, or more briefly,
that 5 is a divisor or factor of 35. Turning the statement around, we say
that 35 is a multiple of 5. In general, let b and d denote any two natural
numbers ; if there is a third natural number q such that b = dq, then d
is said to be a divisor of b, or b a multiple of d. In the example above we
have b = 35 and d = 5, and of course q has the value 7. The letters
d and q were chosen specifically, because they remind us of the words
"divisor" and "quotient. "
1.1 Primes
How many divisors does the number 35 have? The answer is four,
as can be seen by listing all the divisors : 1, 5, 7, 35. The question was
not difficult, because 35 is a relatively small natural number. But now
consider the following question: how many divisors does the number
187 have? This is not so easy to answer, but when we try 1 , 2, 3, etc., it
turns out that again the answer is four, namely 1 , 1 1, 1 7, 1 87. It might
have taken a little effort for the reader to find the divisors 1 1 and 1 7,
but the divisors I and 187 are obvious. Similarly it is apparent that 1 79
has divisors 1 and 1 79, and it turns out that these are the only divisors.
When, as in the case of 1 79, a natural number has exactly two divisors,
such a number is called a prime or a prime number. Another way of
saying this is that a prime is a natural number whose only divisors are
itself and 1. The first few primes in order of size are
2, 3, 5, 7, 1 1 , 1 3, 1 7, 19, 23, 29, 3 1 , 37, 41, 43, 47, . . .
Note that 1 is not listed as a prime. The fact that 1 is not a prime is a
mathematical convention or agreement or, to say this another way, it
is a matter of definition. M athematicians have agreed not to call 1 a
prime. The decision could have been made the other way, to include 1
among the primes. But with 1 excluded, it is possible to state proposi­
tions about primes without making exceptions or qualifications, as will
be shown later.
NATURAL N U M BERS A N D INTEGERS
II
Problem Set 1
[In the problem sets, the starred problems are the more difficult ones.]
I . Decide which of the following statements are true and which are false.
(a) The set 1 , 0,
I is closed under addition.
(b) The set 1 , 0, - I is closed under multiplication.
(c) The set I, 0, - I is closed under subtraction.
(d) The set of positive powers of 2, i.e., the set 21, 2 2, 23, 24, 25, 26, 27,
is closed under multiplication
*(e) The set of pos itive powers of 2 is closed under addition.
-
• • •
,
2. How many divisors does 30 have?
3. How many divisors does 16 have?
4. What is the smallest natural number having exactly
three divisors ?
5. Find all primes between 50 and 100.
* 6.
Prove that if 3 is a divisor of two numbers, it is a divisor of their slim and
their difference. Generalize this and prove that if d is a divisor of two num­
bers bl and b2, then d is a divisor of b1 + b 2 and of bl - b2.
1.2 Unique Factorization
The primes get scarcer as we consider larger and larger natural num­
bers. To illustrate what is meant by this, we point out that there are
1 68 primes
1 35 primes
1 27 primes
1 20 primes
1 19 primes
between 1 and 1000
between 1000 and 2000,
between 2000 and 3000,
between 3000 and 4000,
between 4000 and 5000.
,
Nevertheless, the list of primes is endless ; that is, there are infinitely
many prime numbers. This fact is proved in Appendix A at the end of
the book. The proof does not require any special knowledge and so
the reader can turn to it and read it now if he wishes. We have put this
proof into an appendix because we do not need the result to establish
any other proposition in this book. The proof is given because the
result is interesting in itself.
Now, every natural number, except 1 , either is a prime or can be
factored into primes. For example, consider the natural number 94,860
which obviously is not a prime since
94,860 = 10 X 9486.
12
N U M B E R S : R AT I O N A L A N 0 I R R A T I O N A L
Furthermore, 9486 is divisible by 2, al!'.o by 3, and, in fact, by 9 . Thus
we can write
94,860 = 10 X 2 X 9 X 527
= 2 X 2 X 3 X 3 X 5 X 527.
If 527 were a prime, the above expression would be a factoring of 94,860
into primes. But 527 is not a prime because 527 = 1 7 X 3 1 . Hence we
can write the prime factorization as
94,860 = 2 X 2 X 3 X 3 X 5 X 1 7 X 3 1 .
W e began with the particular number 94,860, but the procedure would
also work no matter what natural number n we started with. For either
n is a prime or it is not. If it is not, it can be factored into two smaller
numbers, say a and b, so that n = abo Each of the numbers a and b, in
turn, either is a prime or can be factored into smaller numbers. Con­
tinuing this process, we finally factor n into primes completely.
The first sentence of the preceding paragraph distinguishes primes
from other natural numbers. It is often desirable, in mathematics, to
make definitions so general that a division into several cases becomes
unnecessary. By "factorization into primes," for instance, we under­
stood the representation of a number, say 1 2, as a product of several
primes, i n this case 2 X 2 X 3. Now let us extend the meaning of
"factorization into primes" so that it will include a single prime.
For example, the prime number 23 then would have a prime factoriza­
tion consisting of the single factor 23. With this extended meaning of
"factoring into primes," our original statement can be replaced by
the sentence : "Every natural number, except 1, can be factored into
primes." Thus we have abbreviated the sentence and eliminated the
necessity of distinguishing prime numbers from other numbers, at
least for the purpose of making a statement about their factorization
into primes.
It is a basic result in mathematics that the factoring of a natural
number into primes can be done in only one way. For example, 94,860
cannot be factored into any primes other than the ones given above.
Of course, the order of the factors can be different ; for instance,
94,860 = 3 X 1 7 X 2 X 5 X 31 X 3 X 2.
But apart from such changes in the order, there is no other way of
factoring 94,860. This result is known as the Unique Factorization
Theorem or the Fundamental Theorem of Arithmetic, which is stated
formally as follows :
N A T U R A L N U M B E R S A N D I N T EGE R S
13
THE FUNDAMENTAL THEOREM OF ARITHMETIC. Every natural number,
other than 1 , can befactored into primes in only one way, apart from the
order of the factors.
This result is proved i n Appendix B . It is a theorem that we shall use
as we proceed with our discussion. The reason for putting the proof
in an appendix is that it is somewhat complicated. However, no ideas
occurring later in the book are used in the proof, so the reader may turn
to Appendix B now if he wishes. Or he may postpone the study of
Appendix B i n order to take easier concepts first, harder concepts later.
The above statement of the Fundamental Theorem of Arithmetic
provides one clue as to why 1 is not included among the prime numbers.
For, if 1 were taken as a prime we then could write, for example,
35 = 5 X 7 = 1 X 5 X 7,
and so 35 (or any other natural number) could be expressed i n more
than one way as a product of primes. Of course the Fundamental
Theorem would still hold but its statement would require more quali­
fying phrases such as "except . . . " or "unless . . . . " Thus by banishing
1 from the list of primes, we can state our results more briefly and
elegantly.
1.3 Integers
The natural numbers 1, 2, 3, 4, . . . are closed under addition and
multiplication, but not under subtraction or division. Closure under
subtraction can be achieved in a set of numbers which is extended to
include zero and the negatives :
0, - 1 , - 2, - 3, -4, . . .
These, taken together with the natural numbers, form the integers or
whole numbers
. . . , - 5, -4, - 3, - 2, - 1, 0, 1 , 2, 3, 4, 5,
The reader is probably familiar with the basic properties
a + b = b + a,
(a + b) + c = a + (b + c),
a + O = O + a = a,
ab = ba,
(ab)c = a(bc),
a · l = 1 · a = a,
a(b + c) = ab + ac .
a·O = O · a = 0,
( -a)( -b) = ab,
14
N U M BERS : RAT I ON A L A N D I R RATIONA L
where a, b, c are any integers. These properties hold for all the number
systems discussed in this book. It is not our intention to discuss the
origins of these particular properties. Such a discussion would lead to
a study of the foundations of the number system (which will be treated
in one of the other books in this series) and away from the topic under
consideration here. Our purpose is to derive various properties of
numbers, especially irrational numbers, taking the foundations for
granted.
The integers, then, are closed under addition, subtraction, and
multiplication. They are not closed under division, because for example,
the result of dividing 2 by 3 is not a whole number and hence leads us
out of the class of i ntegers
Before we define the division of integers, let us examine the other
operations and their results. When we consider addition of integers, we
see not only that the sum of two integers is again an i nteger but also
that there is only one integer which is this sum. For example, the sum
of 3 and - 1 is 2, not 5 and not anything else. We can express this fact by
saying that, given two integers, there exists a unique third integer which
is the sum of the other two. Similarly, for multiplication : given two
integers, there is a unique third integer which is the product of the
other two.
When we discussed the division of natural numbers, we saw that it
was not always true that for any two given natural numbers, say b and
d, there was a third natural number, their quotient, such that b = dq.
However, whenever such a third natural number q does exist, it is clear
that it is the only one, so we did not bother to say that q should be fl
unique natural number such that b = dq. When we define the same
division concepts in the set of integers, however, we must include the
requirement that the quotient be unique. We shall now analyze why
this is necessary.
We must first agree that it is desirable to have only one answer to
each of the questions : How much is 3 - 7? How much is ( - 2) . ( - 3)?
How much is 8 + 4? In other words, we want to obtain a unique result
for our operations. Next, let us see what happens when we consider
division in the set of integers. Again, let b and d be given integers and
define the quotient q to be an integer such that b = dq. For example,
let b = - 12 and d = 3. Clearly, q = -4 because - 1 2 = 3 · ( - 4). An
appropriate q exists and is unique. Next, let b be any integer and let d
be the integer O. We must find a q such that b = 0 . q. If b 'F- 0, t this
equation cannot be solved ; i.e., there is no q for which it is true. If
b = 0, then the equation reads 0 = 0 . q and is satisfied by any integer
t The symbol 'F- means "is not equal to."
N A T U R A L N U M BERS A N D IN T E G E RS
15
other words, if a solution q of b = 0 . q exists at all, it is 1101 unique.
Since unique results of arithmetic operations are important, we must
construct a number system so that the quotient of two integers not only
exists but also is unique. The scheme is simply not to allow division by
zero. We now can state that an integer d is called a divisor of an integer
b if there exists a unique integer q such that b = dq. (Then, by the above
analysis d >= 0.) Or we can say that a non-zero integer d is called a
divisor of b if there exists an integer q such that b = dq. (Since we
barred 0 as a possible divisor, the quotient will automatically be unique.)
In our earlier discussion, we asked the question : how many divisors
does the number 35 have? At that time, the discussion was restricted
to natural numbers and so the answer to this question was four, namely,
I, 5, 7, and 35. If now we interpret the question to mean that the
divisors are to be integers, the answer is eight: ± I, ± 5, ± 7, and ± 35.
q. I n
Problem Set 2
I . Is - 5
a divisor of 35?
2. Is 5 a divisor of - 35 ?
3. I s - 5
a divisor o f -35?
4. Is 3 a divisor of - 35?
5.
Is I a divisor of - 35?
6. Is I a divisor of O ?
7. Is 0 a divisor of I ?
8. Is 1 a divisor of 1 ?
9. Is 0 a divisor of O ?
1 0. Is 1
a
divisor o f every integer?
1 1 . Is 0 a multiple of 35?
1 2. Verify that there are twenty-five primes between 1 and 1 00 and twenty-one
primes between 100 and 200.
1.4 Even and Odd Integers
A n integer is said to be even if it is divisible by 2; otherwise it is said
to be odd. Thus the even integers are
. . . , -8, - 6, -4, - 2, 0, 2, 4, 6, 8, . .
and the odd integers are
. . . , - 7 , - 5, -3, -I, I, 3, 5, 7, . . . .
'
16
N U M BERS : RATION A L A N D I RRATION A L
Since an even integer is divisible by 2, we can write every even integer
in the form 2n, where the symbol n stands for any integer whatsoever.
When a symbol (such as the letter n in our discussion) is permitted to
represent any member of some specified set of objects (the set of
i ntegers in this case), we say that the domain of values of that symbol is
the specified set. I n the case under consideration, we say that every
even integer can be written i n the form 2n, where the domain of n is the
set of integers. For example, the even integers 18, 34, 12, and - 62 are
seen to have the form 2n when n is 9, 17, 6, and - 31 , respectively.
There is no particular reason for using the letter n. Instead of saying
that even integers are integers of the form 2n, we could just as well say
that they are i ntegers of the form 2m, or of the form 2j, or of the form 2k.
If two even i ntegers are added, the result is an even integer. This is
iII ustrated by the examples :
12
14
26
30
22
52
46
- 14
32
- 10
- 46
- 56
However, to prove the general principle that the even integers are closed
under addition requires more than a collection of examples. To present
such proof, we make use of the notation 2n for an even integer and 2m,
say, for another even integer. Then we can write the addition as
2m + 2n = 2(m + n).
The sum 2m + 2n has been written in the form 2(m + n) to exhibit its
divisibility by 2. It would not have been sufficient to write
2n + 2n=4n
because this represents the sum of an even integer and itself. I n other
words, we would have proved that twice an even integer is again even
(in fact, divisible by 4) instead of proving that the sum of any two even
integers is an even integer. Hence we used the notation 2n for one even
integer and 2m for another to indicate that they are not necessarily the
same.
What notation can we use to denote every odd integer? Note that
whenever we add 1 to an even integer we get an odd integer. Thus we
can say that every odd integer can be written in the form 2n + 1 . This
form is not the only one. We could equally well have observed that
whenever we subtract 1 from an even integer we get an odd integer.
Thus we could say that every odd integer can be written in the form
N A T U R A L N U M B E R S A N D I N T EG E RS
17
2n - 1 . For that matter, we can say that every odd integer can be
written in the form 2n + 3, in the form 2n - 3, or in the form 2k - 5,
etc.
Can we say that every odd i nteger can be written in the form 2n2 + 1 ?
If we substitute the integer values
. . . , - 5, - 4, - 3, - 2, - 1 , 0, 1, 2, 3, 4, 5, . . .
for n, we obtain the set of integers
. . . , 5 1 , 33, 19, 9, 3, 1 , 3, 9, 19, 33, 5 1 , . . .
for 2n2 + 1 . Each of these is odd, but they do not constitute al/ odd
integers. For example, the odd integer 5 cannot be written in this form.
Thus it is false that every odd integer can be written in the form 2n2 + I,
but it is true that any i nteger of the form 2n2 + 1 is odd. Similarly, it
is false that every even integer can be written in the form 2k2, where
the domain of k is the set of all integers ; e.g., 6 is not equal to 2k2
no matter which integer is chosen for k. B ut it is true that any integer
of the form 2k2 is even.
The relationship between these statements is the same as that between
the statements "all cats are animals" and "all animals are cats." Clearly,
the first is true and the second is not. This relationship will be discussed
further when we examine statements which include the phrases "if,"
"only if," and "if and only if" (see Sect. 2.3).
Problem Set 3
Which of the following are true, and which are false? (It is u nderstood
that the domain of values for n, m, j,
, i s the set of all i ntegers. )
' "
I . Every odd integer can be expressed in the form
2j - 1 .
(d) 2n2 + 3.
2n + 7.
(e) 2112 + 2n + 1 .
(0 2m - 9.
4n + 1.
2. Every integer of the form (a) above is odd; similarly for (b), (c), (d), (e),
(a)
(b)
(c)
and (f).
3. Every even integer can be expressed in the form
(a)
(b)
(c)
2n + 4.
4n + 2.
2111 - 2.
(d)
(e)
2 - 2111.
n2 + 2.
4. Every integer of the form (a) in the previous problem is even; similarly for
(b), (c), (d), and (e).
18
N U M B E RS: R A T I ON A L A N D I R R A T I ON A L
1.5 Closure Properties
The following two propositions will be of use in a succeeding chapter.
(1) The set of even integers is closed under multiplication .
(2) The set of odd integers is closed under multiplication.
To prove assertion ( 1 ), we must establish that the product of any two
even integers is even. We can represent any two even integers symboli­
cally by 2m and 2n. Multiplying, we obtain
(2m)(2n)
=
4mn
=
2(2I11n).
The product is divisible by 2, and so is even.
To prove assertion (2), we must establish that the product of two odd
integers is odd. Representing the two odd integers by 2m + 1 and
2n + 1, we multiply these to get
(2m + 1 )(2n + 1)
=
4mn + 2m + 2n + 1
=
2(2mn + m + n) + 1 .
Now 2(2mn + m + n) i s even, whatever integers may be substituted for
m and n in this expression. Hence 2(2mn + m + n) + 1 is odd.
Assertions ( 1 ) and (2) could also be proved by an application of the
unique factorization result, but we will not go into details about this
alternative procedure. (The reader may find it challenging to try it by
himself. Let him remember that an integer is even if and only if 2 occurs
in its prime factorization.)
We have concentrated on even and odd integers, i.e., integers of the
form 2m and of the form 2m + 1. Evenness and oddness of integers are
related to divisibility by 2. A nalogous to this, we can consider the class
of integers divisible by 3, namely
. . . , - 1 2, - 9, - 6, - 3, 0, 3, 6, 9, 12,
.
. .
.
These are the multiples of 3. They can also be described as the class
of integers of the form 3n. The integers of the form 311 + 1 are
. . . , - 1 1 , - 8, - 5, - 2, 1 , 4, 7, 10, 1 3 , . . . ,
and the integers of the form 3n + 2 are
.
.
.
,-
10
,-
7
, -
4
,-
1 2 5 8
,
,
,
.
11
,
14
, .
.
.
.
These three lists of integers include all integers; thus we can say that
any integer is of exactly one of the forms 3n, 3n + 1, or 311 + 2.
NATURAL N U M BERS A N D I NTEGERS
1.6
A
19
Remark on the Nature of Proof
We �aid earlier that in order to prove that the even integers are closed
under addition, i.e., that the sum of any two even integers is even, it
would not suffice to examine only a few specific examples such as
12 + 14 = 26 . Since there are infinitely many even integers, we cannot
check all cases of sums of specific pairs of even integers. So it is necessary
to turn to some kind of algebraic symbolism; for example, the symbol
2n, which can be used to express any even integer, enabled us to prove
the closure of the set of all even integers under multiplication.
However, to prove a negative proposition such as "The odd integers
are not closed under addition," we do not have to use any general
algebraic symbols like 2m + 1 . The reason for this is that such a nega­
tive assertion can be established by a single example. To prove any
statement which asserts that not all members of a set have a certain
property, it clearly suffices to find one single member which does not
have that property. To prove that not all boys have brown eyes, we
need only find a blue-eyed or a hazel-eyed boy. To prove that not all
sums of two odd integers are odd, observe that 3 + 5 = 8, and this
single case of the addition of two odd integers giving an even sum is
sufficient proof. However, if we want to prove that the sum of any two
odd integers is an even integer, it does not suffice to write 3 + 5 = 8.
Even if we write many cases, 7 + II = 18, 5 + 53 = 5 8, etc., we
would not have a correct mathematical proof of the proposition.
Another example of a negative proposition is: "Not every prime
number is odd." To prove this, we need merely point out that the even
number 2 is a prime.
Problem Set 4
(The first three problems involve negative propositions and so can be solved
by giving a s ingle numerical example.)
I . Prove that the odd integers are not closed under subtraction.
2. Prove that the integers of the form
3n + 1 are not closed under addition.
3. Prove that the integers of the form 3n + 2 are not closed under multiplica­
tion.
4. Prove that the sum of any two odd integers is an even integer.
5. Prove that the following sets are closed under the indicated operation :
(a) the integers of the form 3n + I, under multiplication ;
(b) the integers of the form 3n, under addition ;
(c) the integers of the form 3n, under multiplication.
6. Decide which of the following sets are closed under the indicated operation,
and give a proof in each case :
20
N U M B E R S : R ATI O N A L A N D I R R A T I O N A L
(a)
(b)
(c)
(d)
(e)
(0
(g)
the integers of the form 6n + 3, under addition;
the integers of the form 6n + 3, under multiplication;
the integers of the form 6n, under addition;
the integers of the form 6n + I , under subtraction ;
the integers of the form 6n + I , under multiplication;
the integers of the form 3n, under multiplication;
the integers not of the form 3n, under multiplication.
C H A PT E R T W O
Rational Numbers
2.1 Definition of Rational Numbers
We have seen that the natural numbers 1 , 2, 3, 4, 5,
under addition and mUltiplication. and that the integers
are closed
. . . , - 5, - 4, - 3, - 2. - 1, 0, 1 , 2. 3, 4, 5,
are closed under addition, multipl ication, and suhtraction. However,
neither of these sets is closed under division, because division of integers
can produce fractions like 4/3, 7/6, - 2/5, etc. The entire collection of
such fractions constitutes the rational numbers. Thus a rational number
(or a rational fraction) is a number which can be put in the (orm aid,
where a and d are integers, and d is not zero. We have several remarks
to make about this definition :
(1) We have required that d be different from zero. This requirement,
expressed mathematically as d 'F- 0, is necessary because d is in effect
a divisor. Consider the instances :
Case (a)
a
21,
d = 7,
3
a 21
- = - = - = 3''
d
7
I
Case (b)
a = 25,
d = 7,
4
a 25
d=7=3-1'
=
In case (a), d is a divisor in the sense of the preceding chapter ; that is,
7 is an exact divisor of 2 1 . In ca�e (b), d is still a divisor, but in a
different sense, because 7 is not an exact divisor of 25. But if we call
25 the dividend, and 7 the divisor, we get a quotient 3 and a remainder
4. Thus we are using the word divisor, in a more general sense, to
cover a wider variety of cases than in Chapter 1 . However, the divisor
21
22
N U M B E R S : R A T IO N A L A N D I R R A T IO N A L
concept of Chapter 1 remains applicable in such instances as case (a)
above ; hence, as in Chapter 1 , we must exclude d = O.
(2) Note that while the terms rational number and rational fraction
are synonymous, the word (ractiol/ alone is used to denote any
algebraic expression with a n umerator and a denominator, such as
V3
2'
17 ,
x
-
x2
x2
or
_
_
y2
y2
(3) The definition of rational number included the words "a number
which can be put in the form aid, where a and d are integers, and
d ;t. 0." Why is it not enough to say "a n umber of the form aid,
where a and d are integers, d ;t. O"? The reason is that there are
infinitely many ways to express a given fraction (for example, 2/3
also can be written as 4/6, 6/9, . . , or 271/371, or 2 v3/3 v3, or
- 10/ - 1 5, j ust to mention a few) and we do not want our definition
of rational number to depend on the particular way in which some­
body chooses to write it. A fraction is so defined that its value does
not change if its numerator and denominator are both multiplied by
the same quantity ; but we cannot always tell, just by looking at a
given fraction, whether or not it is rational. Consider, for example,
the numbers
.
v 12
v3
and
neither of which, as written, is in the form aid, where a and dare
integers. However, we can perform certain arithmetic manipulations
on the first fraction and obtain
We thus arrive at a number equal to the given fraction but of the
specified form : a = 2, d = 1 . Thus we see that V 12/ v3 is rational,
but it would not have q ualified had the definition stipulated that the
number be in the right form to start with. In the case of V l S/V3,
the manipulations
v IS
v5 . v 3
v3
v3
v5
R A T I O N A L N U M BE R S
23
yield the number V5. We shall learn. in the following chapters, that
..,Is cannot be expressed as a ratio of integers and hence is irrational.
(4) Note that every integer is a rational number. We have j ust seen
that this is so in the case of the integer 2. In general , the integers can
be written in the form
- 5'
1
-4
-1
--
-3
- 2'
-1
--'
1
l' l' l' l' l' l'
0
2
3
4
5
where each is given the denominator 1 .
Problem Set 5
I. Prove that the integer 2 can be written in rational form aid (with integers
a
and d) in infinitely many ways.
2. Prove that the rational number 1/3
can be written in rational form aid in
infinitely many ways.
3. Prove that the integer 0
many ways.
can be written in rational form aid in infinitely
4. Prove that every rational number has infinitely many representations in
rational form.
5. Definition. Let k be any number; then the reciprocal of k is another number,
say I, such that k . I = I.
This definition has the consequence that all numbers except 0 have recipro­
cals. Given k ;t. 0, by definition its reciprocal I satisfies the equation
k.I
I ; hence
=
I
=
!
k
which is meaningful only for k � O . Prove that the reciprocal of any
rational number (except zero) is rational.
2.2 Terminating and Non-terminating Decimals
T here is a n ot he r re prese ntatio n of the rational number 1 /2 which is
different from the forms 2/4, 3/6, 4/8, etc., namely as a decimal , 0.5.
The decimal re presentatio n s of some fractions a re ter m i nati n g, or
finite ; for example,
1
2 = 0.5 ,
"5 = 0.4,
2
1
80
=
0.0125.
24
N U MBERS : RATION A L AND I RRATIONAL
Other fractions, however, have non-terminating, or infinite, decimal
representations ; for example,
1
6 = 0. 1 6666· . . ,
1
3 = 0.33333 . . . ,
5
IT
= 0.454545 . . .
These infinite decimals can be obtained from the fractions by dividing
the denominator into the numerator. In the case 5/1 1 , for example,
we divide 1 1 into 5.000 . . . and obtain the result 0.454545 . . . .
Which rational fractions alb have terminating decimal representa­
tions? Before we answer this question in general, let us examine an
example, say the terminating decimal 0.8625. We know that
0.8625 =
�,
8
1
and that any terminating decimal can be written as a rational fraction
with a denominator which is 10, 100, 1000, or some other power of 10.
If the fraction on the right is reduced to lowest terms, we get
0. 8625 =
8625
69
= .
10000
80
The denominator 80 was obtained by dividing 10,000 by 1 25, the
greatest common factor of 10,000 and 8625. Now the integer 80, like
10,000, has only the two prime factors 2 and 5 in its complete prime
factorization. If we had started with any terminating decimal whatever,
instead of 0.8625, the corresponding rational fraction form alb in lowest
termst would have the same property. That is, the denominator b could
have 2 and 5 as prime factors, but no others, because b is always a
factor of some power of 10, and 10 = 2· 5. This turns out to be the
deciding issue, and we shall prove the general statement :
A rational fraction alb in lowest terms has a terminating decimal
expansion if and only if the integer b has no prime {actors other than
2 and 5.
It should be understood that b does not have to have 2 and 5 as prime
factors ; it may have only one or perhaps neither as prime factors :
1
= 0.04,
25
1
= 0.0625,
16
7
1
= 7.0,
t A rational number alb is in lowest terms if a and b have no common divisor d
greater than 1.
RATIONAL N UMBERS
25
where b has the values 25, 1 6, and 1. The important notion is that b
must not have any prime factors other than 2 and 5.
Note that the above proposition contains the words if and only if.
So far, all we have proved is the only If part because we showed that
there would be a terminating expansion only if b is divisible by no primes
other than 2 and 5. (In other words, if b is divisible by primes other
than 2 and 5, then the fraction alb in lowest terms will not have a
terminating decimal expansion.)
The if part of the proposition states : if the integer b has no prime
factors other than 2 and 5, then the rational fraction alb in lowest terms
has a terminating decimal expansion. To prove the if part, we must
begin with any rational fraction alb in lowest terms, assume that b has
at most the prime factors 2 and 5, and prove that the corresponding
decimal expression is of the terminating type. Let us first consider an
example, say
9741
a 9741
'
b = 3200 = 27 . 52
To convert this into a decimal, we merely change it into a fraction
whose denominator is a power of 10. This can be achieved if we multiply
both the numerator and denominator by 55 :
30440625
9741 . 55
9741
= 3 . 0440625 .
=
=
27 . 52
27 . 57
107
This argument can be generalized from this special case to any instance
whatever in the following way : Suppose that b is of the form 2111 . 5",
where m and n are positive integers or zero. Now, either n is less than
or equal to m (written n � m) or n is greater than m (written n > m).
When n � m, we multiply both the numerator and denominator of the
fraction by 5",-11 :
a
a
- = 111
h
2 . 5"
--
-
a · 5'"-"
111
2 . 5" . 5"'-"
a · 5111-11
2"' · 5'"
a · 5 ",-1I
----wm-'
Since m
n is positive or zero, 5'''-11 is an integer, and so a · 5111-11 is
also an integer, say c. Hence we can write
-
and since division of the integer c by 10m merely requires that we insert
the decimal point at the correct place, we get a terminating decimal.
26
N U M BERS : RATION A L A N D I R RAT I O N A L
On the other hand, if n > m, we would multiply the numerator and
denominator of alb by 2/1-111:
a
-
h
=
a
--
2m S"
=
•
a 2"-111
2" . S"
a . 2/1-111
111
2 . S/I . 2/1-111
.
a . 2"-111
1 0"
Writing d for the integer a 2/1-111, we get
.
a
d
b = 10" '
Thus we have a terminating decimal as before.
Problem Set
I. Express
(a)
6
the fol lowing fractions as terminating decimals :
�,
(b)
f60'
(c)
��� ,
(d)
6�5'
(e)
�;;,
(f)
;��6 '
2.3 The Many Ways of Stating and Proving Propositions
We have used the phrase If and on�F if without precisely defining it.
So at this point we pause in our discussion of rational numbers to
explain briefly some of the language used in making mathematical
statements, and also the relation of this language to the underlying
logic. There are two basic kinds of assertions or propositions in
mathematics:
If A then B.
I f A then B, and conversely.
We take these up in turn.
When we state "if m and n are even integers, then mn is even," as
we did in Section I .S, we have an "if A then B" kind of assertion. Now,
such an assertion may be stated in many ways, as illustrated by the
following list :
Ways of Stating "If A then B"
[1]
[2]
[3]
[4]
If A is true then B is true.
If A holds then B holds.
A implies B.
B is implied by A .
RATION A L
N U M BERS
27
B follows from A .
A is a sufficient condition for B.
B is a necessary condition for A .
B i s true provided A i s true.
B is true if A is true.
A is true only if B is true.
It is impossible to have A true and B false
at the same time.
[12] If B is false then A is false.
[5]
[6]
[7]
[8]
[9]
[10]
[11]
This list includes only the most common forms and is not complete,
since there is virtually no limit to the number of possible forms of the
statement. Some items, [6] and [7] for example, are not used in this
book, but are included for completeness. All but [12] can be regarded
as definitions of such terms as "implies," "necessary condition,"
"sufficient condition," and "only if."
Consider [10], for example, which defines the technical use in mathe­
matics of the term "only if." Replacing the symbols A and B with the
assertions about m and n of our previous discussion, we conclude that
the following propositions both say the same thing.
"If the integers m and
n
"The integers m and
are even only if the integer mn is even."
n
are even, then the integer mn is even."
If the reader feels that they do not say the same thing, his feeling arises
from some day-to-day usage of the word "only" to which he is accus­
tomed. In this case, he should recognize a distinction between the
technical language of mathematics and the everyday use of English.
While these languages have much in common, there are pointed
differences, as in the example under discussion. (After a person becomes
skilled in the mathematical use of language, he can, if he chooses, use
it as the form of his day-to-day speech. However, if he so chooses, he
will be regarded by the man-in-the-street as pedantic, affected, or, at
least, stuffy.)
What we have said thus far about the list of ways of saying "if A
then B" is that forms [1] to [1 1] are based on nothing more than agree­
ments about the way we use language in mathematics. Form [ 1 2]
involves not only a new wording but also a fundamental axiom of logic.
The fact that [12] says the same thing as "if A then B" has its basis in
logic, and not simply in a different arrangement of words. The axiom
of logic referred to (known as the law of the excluded middle) states that
either A is true or A is false, where by A we mean any statement capable
of analysis. In essence, the axiom excludes any middle ground between
28
N U MBERS : RATION A L A N D I RRATIONAL
the truth and the falsity of A . Let us take this axiom for granted, and
then prove that forms [1] and [12] say the same thing.
To do this, we must prove that [1] implies [12], and conversely, that
[ 1 2] implies [1]. First, we assume [1] and then consider [12] :
"If B is false then A is false."
Is it possible that this conclusion is wrong, that it should be "A is
true" ? If this were so, then by use of [1] we would conclude that B is
true, but this contradicts the hypothesis in [12]. Therefore, the con­
clusion "A is false" is right.
Conversely, let us assume [ 1 2] and prove that [1] follows :
"If A is true then B is true."
We ask whether this conclusion is wrong ; should it be "B is false" ? If
this were so, then by use of [ 1 2] we would conclude that A is false, but
this contradicts the hypothesis in [1]. Therefore " B is true" is the correct
conclusion.
Forms [ 1 1] and [ 1 2] give the clue to the nature of indirect proof.
Suppose we want to prove the assertion "if A then B." A direct proof
is one in which the statement A is taken for granted, or assumed, and
then the statement B is deduced. But if we examine form [1 1], we see
that we can give a proof by assuming both the truth of A and the falsity
of B, and then deducing a contradiction. This is a proof by contradic­
tion, one of the methods of indirect proof. This type of proof can be
spotted by noting the assumptions formulated in the proof; one is
usually asked to assume first that the statement which is actually to be
proved is false. Indirect proofs can also be spotted by the kind of
language that occurs at the end of the proof, such as " . . . and so we
have a contradiction and the theorem is proved."
Another common type of indirect proof is suggested by [ 12]. Thus
to prove "if A then B," we can assume that B is false and then deduce
that A is false. The three types of proof that we have identified are :
Assume A , deduce B. (direct proof)
Assume A true and B false, deduce a contradiction . (a form of
indirect proof)
Assume B false, deduce that A is false. (another form of indirect
proof)
Now, a curious thing about the way in which mathematics books are
written (including the present one) is that these three types of proof are
used freely, but often without any clear indication as to which type is
being employed at any particular time ! In effect, the reader is expected
to solve a little riddle by identifying for his own thought processes the
RATIONA L N U M BERS
29
type of proof being used. The riddle is not a difficult one, however,
and the reader usually can spot the assumptions made by the writer at
the start of the proof.
Next, let us consider the second kind of mathematical proposition :
"If A then B, and conversely,"
which is mentioned at the beginning of this section. The words and
conversely have the significance "if B then A," and this is the converse
of "if A then B." The reader is probably aware that a statement and its
converse are two different things. One may be true and the other false,
both may be true, or both may be false, depending on the circumstances.
For example, the statement "if m and n are even, then mn is even" is
true, whereas the converse "if mn is even, then m and n are even" is false.
We now parallel our earlier list and indicate various ways of stating
"if A then B, and conversely" :
If B then A , and conversely.
A is true if and only if B is true.
B is true if and only if A is true.
A is false if and only if B is false.
B is false if and only if A is false.
A implies B, and conversely.
B implies A , and conversely.
A is a necessary and sufficient condition for B.
B is a necessary and sufficient condition for A .
A and B are equivalent statements.
All of these statements say the same thing.
But now let us note the wide variety of methods of proof available
for est ablishing "if A then B, and conversely." As we saw earlier, there
are three basic approaches to the proof of "if A then B." Similarly there
are three possible methods of proof for "if B then A." Since any one of
the first three may be combined with any one of the second three, there
are nine possible organizations of the proof of "if A then B, and con­
versely." Perhaps the most common pattern is the direct proof each
way : i.e.,
(1) Assume A, deduce B.
(2) Assume B, deduce A .
Another common pattern i s this :
( 1 ) Assume A, deduce B.
(2) Assume A false, deduce that B is false.
In
somewhat complex proofs, these patterns are often combined. A
30
NUM B E R S :
RATI ONA L
AND
I R RATI ONA L
proof of "if A then F" may be built by means of a chain of statements :
"if A then B," "if B then C," "if C then D," "if D then E," "if E then
F. " In this case, each statement implies the next. Now, if each statement
and its converse can be established by one of the above patterns, then
we also have "if F then E," "if E then D," "if D then C," "if C then B,"
"if B then A," so that the converse "if F then A" of the original state­
ment also holds. When an author says "the converse can be proved by
reversing the steps," this is what he means.
All these patterns can be found in mathematical books, and as we
said before, the writer will often launch forth on the proof of a theorem
with no clear declaration of which pattern he is following. The writer
expects the reader to figure out for himself the nature of the proof
technique.
Problem Set
I. Prove that the statement "if mn is
7
even then
,
III
and n are even" is false.
2. Which of the following assertions are true, and which are false? The rational
fraction alb in lowest terms has a termi nati ng decimal representation :
(a) if and only if b is divisible by no prime other than 2 ;
(b) i f b i s divisible by no prime other than 2 ;
(c) only i f b i s divisible by no prime other than 2.
(d) if and only if b is not divisible by 3 ;
(e) if b is not divisible by 3 ;
(f) only if b is not divisible by 3.
3. Which of the following statements are true, and which are false? The
rational fraction alb has a terminating decimal expansion :
(a) if and only if b has no prime factors other than 2 and 5 ;
(b) if b has no prime factors other than 2 and 5 ;
(c) only i f b has no prime factors other than 2 and 5.
Suggestion. Observe that it has not been specified that alh is in lowest
terms.
A recent book on algebra t empl oys the following statement as an axiom :
"ab 0 only if a 0 or b 0." Rewrite this in the form "if A then B."
5. (a) Prove that if {:J (beta) is a rational number, then {:J2 is also rational .
4.
=
=
=
(b) Does this amount to proving that if {32 is irrational, then {:J is irrational?
2.4 Periodic Decimals
We now return to the topic of rational numbers. We have separated
rational fractions into two types, i.e., those with terminating decimals
t w. W. Sawyer, A
Concrete Approach to Abstract Algebra, p. 30.
RATIONAL
N U M BERS
31
and those with infinite decimals. We now can establish that each such
infinite decimal has a repeating pattern such as
5
= 0. 454545 . . .
IT
3097
= 0.3 12!Q!Qg . . . .
9900
and
For convenience we will use the standard notation to indicate a periodic
decimal, namely by the use of a bar over the repeating part :
3 = 0.3,
5
= 0.45,
IT
1
-
-
1
6
6 = 0. 1 ,
-
etc.
The reason for the repetitive pattern of digits can be seen from a con­
sideration of the standard method of converting a fraction, 2/7 for
example, into decimal form :
. 2g57 14
7)2.000000
14
60
56
40
35
50
::; = 0. 2857 14
2
-­
49
10
7
30
2g
2
In the division process, the successive remainders are 6, 4, 5, I, 3, 2 .
When the remainder 2 i s reached, the cycle i s complete and we have a
recurrence of the division of 7 into 20. The remainders are all less than
the divisor 7, so there must be a recurrence since there are only six
possible remainders . (The remainder 0 is ruled out of consideration
because we are not examining terminating decimals.)
In the above example, the recurrence happened when the division of
7 into 20 turned up for a second time. Now, 7 i nto 20 was the first step
in the whole division process. It need not happen that the first step is
the one that recurs. Consider, for example, the conversion of 209/700
into decimal form :
32
N U M B ER S :
RATIONAL AN D
.29857142
7(0)209.000000
140 0
69 00
63 00
6 000
5 600
4000
3500
5000
4900
1000
700
3000
2800
2000
1400
600
I RRATIONAL
;� = 0.29857142
The recurrence here happens when we reach the remainder 600, which
had occurred several steps earlier. With 700 as the divisor, we know
that the possible remainders are the numbers 1, 2, 3, . . . , 699. Thus we
are sure of a repetition of a remainder although we might have to
follow through q uite a few steps to reach the repetition.
The general case, alb, can be argued in a similar manner. For, when
the integer b is divided into the integer a, the only possible remainders
are 1 , 2, 3, . . . , b
2, b
1 , and so a recurrence of the division process
is certain. When the division process recurs, a cycle is started and the
result is a periodic decimal.
What we have proved so far is half of the following proposition :
-
-
Any rational fraction a/b is expressible as a terminating decimal or an
infinite periodic decimal ; conversely, any decimal expansion which is
either terminating or infinite periodic is equal to some rational number.
The converse deals with two kinds of decimals, terminating and infinite
periodic. The terminating decimals have already been discussed and
we have seen that they represent rational numbers. Let us now turn to
the infinite periodic decimals. We first shall show, using a method
which can be generalized to cover all cases, that a particular infinite
periodic decimal is rational. After treating a particular case we shall
apply the same method to any periodic decimal.
RATION A L
N U M BERS
33
Consider the infinite periodic decimal
x =
28. 1 23456
or
x = 28. 1 23456456 · . . .
We shall multiply it first by one number and then by another ; these
numbers will be so chosen that, when we take the difference of the two
products, the infinite periodic part will have been subtracted out. I n
the example, the numbers 10 6 and 1 0 3 will serve this purpose because
10 6 • x
and
=
103 • x =
so that the difference 106 . X
-
999000x
=
28 1 23456 .456
28 1 23 .456 ,
103 . X is
28095333 .
Therefore
28095333
x = 999000 '
which exhibits the fact that x is a rational n umber.
In generalizing this method we shall show that the numbers 103 and
106 were not "pulled out of a hat" but were chosen systematically. We
shall omit the integer part of the decimal (i.e., the part that corresponds
to 28 in the above example) because it plays no decisive role in the
process. Thus we may write any repeating decimal (without integer
part) in the formt
where a(, a2, . . . , as represent the s consecutive digits in the non­
repeating part and b(, b2,
hI represent the t digits in the repeating
.
•
• ,
t Note that the notation ala2 . . . ashlb2 . . . bl used here is not the usual algebraic
notation and does not denote the product of the numbers a!. a2, . . . , bl; in this
proof it means the integer whose digits are aI, a2, . . . , bl• Furthermore, the symbols
1 , 2, . . . , s in the notation aI, a2, . . . , as are called "subscripts" and have no sig­
nificance except as identification tags ; without subscripts we would soon run out of
letters.
34
N U M B E RS :
R AT I ON A L A N D
I R R AT I ON A L
part. (In the above example, S = 3 , t = 3 ; a l = 1, a2 = 2, a3 = 3,
b l = 4, b2 = 5, and b 3 = 6.) If we multiply x first by lOS+t, then by
l OS, and subtract, we obtain
IOS+/ X = ala2 '
lOS . x = ala2
•
•
•
•
•
•
asb1b2
•
•
•
as + .b1b2
bl + .b1b2
•
•
•
•
•
•
hI>
bl;
and
so that
which is of t he integer-divided-by-integer form. Hence it is rational as
we set out to prove.
Problem Set 8
1. Find rational numbers equal to the following decimals :
(b) 5.6666 · . .
(c) 0.3743
(a) 0.1 1 1 . . .
(d) 0.9987
(e) 0.0001
(0
0.9
2.S Every Terminating Decimal Can Be Written As a Periodic Decimal
What we have established i n this chapter is that some rational num­
bers can be expressed as terminating decimals, whereas other rational
n umbers turn out to be infinite or non-terminating decimals. Curiously
enough, every terminating decimal (except zero) can be expressed in a
non-terminating form. Of course this can be done i n a very obvious
way when we write 6.8 as 6.8000 . , i.e., with an infinite succession
of zeros. B ut apart from this obvious process of changing a terminating
decimal into a non-terminating one by appending a whole string of
zeros, there is another way that is a little surprising. Let us begin with
the well-known decimal expansion for 1 /3 :
.
.
3' = 0.33333 · ·
1
'
.
If we multiply both sides of this equation by 3, we get the strange­
looking result
(1)
1 = 0.99999 · · · .
RATIONAL
N U M BERS
35
Thus we have equality between the terminating decimal 1, or 1 .0, and
the nonterminating decimal 0.99999 . . . .
Let us look at equation (1) in another way. Suppose we denote the
infinite decimal 0.99999 . . . by x; that is,
x = 0.99999 · · · .
(2)
M ultiplying by 10, we get
lOx = 9.99999 . . . = 9 + 0.99999 . . . .
Subtracting eq. (2) from this, we obtain
9x = 9
or
x = 1.
Thus we have proved eq. ( 1 ) by a different approach from that used
initially.
Now, upon division by 10, 100, 1000, 10,000 etc., eq. ( 1 ) yields the
entire succession of results
0. 1 = 0.099999 . . .
om
= 0.0099999 . . .
0.001 = 0.00099999 . . .
(3)
0.000 1 = 0.00009999
.
.
.
•
etc.
These results can be used to convert any terminating decimal into a
non-terminating form. For example, we can write
6.8 = 6.7 + 0. 1
=
6.7 + 0.099999 · . . = 6.799999 · . . .
Some further examples are :
0.43 = 0.42 + om
=
0.758 = 0.757 + 0.001
0.42 + 0.0099999 . . . = 0.4299999 . . . ;
=
0.757 + 0.00099999 · . .
=
0.75799999 · . . ;
0. 102 = 0. 101 + 0.001 = 0. 101 + 0.00099999 · . . = 0. 10199999 · . . ;
6.8 1 = 6.8 + 0.01 = 6.8 + 0.0099999 · . .
=
6.8099999 · . . .
This device enables us to write any terminating decimal in non-termi­
nating form. Conversely, eqs. ( 1 ) and (3) can be used to convert any
36
N U MBER S :
RATI O NAL AN D
IR RATIONA L
decimal that has an infinite succession of nines into a terminating
decimal :
0.4699999 · . . = 0.46 + 0.0099999 . . . = 0.46 + om = 0.47,
1 8.099999 · .
.
=
1 8 . + 0.099999 . . = 1 8. + 0. 1
.
=
18.1.
The question o f how many representations o f a given number there
are by decimals involves a matter of interpretation. For, in addition to
writing 0.43 as 0.42999 . . " we can also write this number in the forms
0.430,
0.4300,
0.43000,
0.430000,
....
These, however, are such trivial variations on 0.43 itself that we do not
count them as essentially different representations. When we refer to
the infinite decimal form of a n umber such as 0.43, we mean 0.42999 . .
and not 0.43000 . . .
.
'
Problem Set 9
I.
Write each of the following as a terminating decimal :
(d) 9.999 · . .
(c) 4.79999 · . .
(b) 0.299999 · . .
(a) 0. 1 1999 . . .
2. Write each of the following as a non-terminating decimal :
(b) 0.0099
(c) \ 3
(a) 0.73
3. Which rational numbers alb have two essentially different decimal repre­
sentations?
4. Which rational numbers alb have three essentially different decimal repre­
sentations?
2.6 A Summary
We have distinguished two types of rational numbers a/b, those in
which the integer b has no factors other than 2 and 5, and all others.
(It is presumed that a/b is in lowest terms.) Those of the first type can
be written both as finite and infinite decimals ; for example,
1
2 = 0.5 = 0.499999 . . . .
The numbers of the second type can be written only in infinite decimal
form ; for example,
"3
1
=
0.33333 · . . .
RATI O N A L N U M B E R S
37
These representations are the only possible ones in the sense that 1/2
and 1/3 cannot be expressed in any other decimal form excluding, of
course, such trivialities as 0.500. We shall explain in the next chapter
why this is so.
The emphasis has been on rational numbers and their decimal repre­
sentations. Turning the matter around, let us reflect on decimal repre­
sentations for a moment. All the infinite decimal expansions in this
chapter have been periodic. What about non-periodic decimals such as
q
=
0. 101 001 000 100 001 000 001 000 000 1 · · . . .
formed by a series of ones separated by zeros, first one zero, then two
zeros, then three zeros, and so on? What kind of a number, if it is a
number, is q ? From our studies in the present chapter we know that q
is not a rational number. I n the next chapter we shall broaden our
inquiry to include such numbers as q.
C H A PT E R T H R E E
Real Numbers
3.1 The Geometric Viewpoint
When coordinates are introduced in geometry, one straight line is
designated as the x-axis, say, and this axis is graduated so that each
point is associated with a number. This is done by taking two arbitrary
(but distinct) points on the line as the positions for 0 and 1 such that
the distance between these two points becomes the unit of length, or
the unit length. It is conventional (Fig. 5) to take the I-point to the
-2
I
1
I
-
o
I
Figure 5
1
I
2
I
right of the O-point, so that the points to the left of the O-point are
associated with negative numbers. The O-point is called the origin. The
point belonging to the number 7, for example, is seven units of length
to the right of the origin. The point belonging to - 7 is seven units to
the left of the origin. In this way a number is attached to each point,
the number being the distance from the point to the origin either with
a plus sign if the point is to the right of the origin or with a minus sign
if it is to the left. As shown in Fig. 6, rational numbers like -4/3, 1/2,
and 2.3 are readily located by their relation to the unit length.
-2
o
1
2'
I
1 ./2
2
2.3
Figure 6
The symbol vi denotes a number which, when multiplied by itself,
yields 2 ; that is, vi . vi = 2. To see the geometric meaning of vi,
we consider a unit square as shown in Fig. 7, and we find from the
Pythagorean Theorem that the square of the length of its diagonal is 2.
38
REA L
N U M BERS
39
1
{21
Figure 7. A square with sides of length I
Hence, we denote the length of the diagonal by ,,/2 and associate the
number v2 with that point on the line whose distance from the origin
is equal to the length of the diagonal of our unit square.
Since each point on the axis lies at some distance from the origin,
it is intuitively clear that there is a number associated with each such
point. By the real numbers we mean the collection of all the numbers
associated with all the points. Every rational number is included be­
cause there is an appropriate distance from the origin for each rational
number. Thus we can say that the rational numbers form a subclass of
the real numbers.
H owever, there are real numbers which are not rational. The number
\/2 is not rational, as we shall prove later in this chapter. Any real
number, such as vi, which is not rational is said to be irrational .
Because of the way the definitions have been made, any real number is
either rational or irrational. The straight line, or axis, with a number
attached to each point in the manner described above, is called the
real line. The points on this line are referred to as rational or irrational
points depending on whether the corresponding numbers are rational
or irrational.
Note that the above definition of an irrational number amounts to
this : any real number which cannot be expressed as the ratio alb of two
integers is called an irrational number.
3.2
Decimal Representations
The number 1/ is easily located on the real line at a point of tri­
3
section between the zero- and unit-points (Fig. 8). Now consider the
0
I
"3
1
I
Figure 8
1
decimal representation of 1/ :
1
3"
3
.
3 + 3
3
=0
.33333 · = 1 0 + 1 00 1000 + . . . .
.
40
N U MBERS : RATIO N A L A N D I R RATIONA L
This equation expresses 1 /3 as a sum of infinitely many terms. Even
t hough there is no end to the number of terms, the sum has a definite
value, i.e., 1/3. If we locate the points associated with
0.3,
0.33,
0.333,
0.3333, . . .
on the real line, we get a sequence of points which converge on the
point 1/3. This is illustrated in Fig. 9, where the unit of length has been
I
o
I
"3
II
.5
0 30 033
Figure 9
magnified. In the same way, any infinite decimal belongs to a particular
point on the real line. In the case of the infinite decimal 0.99999 · . . ,
the point it represents is converged on by the points associated with
0.9,
0.99, 0.999, 0.9999,
0.99999,
etc.
As shown in Fig. 10, these points are converging on the point 1 , in
09
I
o
0 99
II
Figure 10
accordance with the equation 1 = 0.99999 · . . of the preceding chapter.
Now, when we turn to the number
q =
0. 101 001 000 100 001 000 001 000 000 1 · .
which was used as a previous example, we find that this number also
belongs to a particular point on the real line. This point can be thought
of as being converged on by the following chain of points :
0. 1 ,
0 101,
.
0. 101 001,
0. 101 001 000 1 ,
0. 101 001 000 100 001, etc.
REAL
N U M BERS
41
Since q is a non-periodic decimal, it is an irrational n umber, and the
corresponding point is an irrational point.
This suggests another way of interpreting real numbers. The real
numbers are the collection of all decimal expansions, finite or infinite,
such as
1 7.34,
2. 1 76,
- 6.037 222 22 . . .
,
q
=
. 101 001 000 1 . . . .
According to our studies i n the preceding chapter, we can separate
these decimal expansions into rational and irrational numbers. The
rational numbers are those decimals which are either terminating or
periodic ; the irrational numbers are those which are non-periodic, such
as the number q above. M oreover, since we have seen that every
terminating decimal (or every decimal like 0.43000 . . . with an infinite
succession of zeros) can also be written as a genuinely infinite periodic
decimal, we can agree to write, within this section, all rational numbers
as infinite periodic decimals. (By such an agreement we would, for
example, write 0.43 in the form 0.42999 . . . ; this may seem awkward,
but it will simplify the discussion below.)
We shall now show that real numbers have a unique representation as
infinite decimals. That is to say, two infinite decimals are equal only if
they are identical, digit by digit.
Why is the infinite decimal representation unique? We answer this
question as follows : consider two numbers with different infinite
decimal representations. Since the representations are different, there
is at least one digit wherein this difference can be actually observed ;
for example,
a
=
1 7.92341 6 · · . ,
b
=
1 7.9234 1 5 · . . .
The infinite succession of digits that follow 6 in the number a may be
any collection that the reader chooses to imagine, except an infinite
succession of zeros. A similar remark applies to the number b. Now,
the fact that an infinite succession of zeros is excluded tells us that a
is definitely larger than 1 7.92341 6, which in symbols is expressed as
a > 1 7.923416.
On the other hand, b is at most 1 7.92341 6, because we can have
b = 1 7.923416 only when the succession of digits in b following the
42
N UM BERS :
RATIONA L
AN D
I RRATION A L
"5" are all nines, i.e., when b = 1 7.9234 1 59 . The statement that b is at
most 1 7.92341 6 is written symbolically as
b � 1 7.92341 6
These inequalities for a and
h
1 7.92341 6 ;:;; b.
or
state that
a > 1 7.92341 6 ;:;; b
and hence a > b. We have concluded, then, that a is greater than h,
and, of course, this rules out the possibility of equality. Our argument
has been applied to a special case of two particular numbers a and b
but the reasoning generalizes at once to any pair of numbers having
different infinite decimal representations.
3.3 The Irrationality of V2
We now give the traditional indirect proof that v2 is irrational,
and in the next chapter we shall give yet another proof by means of a
much more general approach.
In Chapter 1 we showed that the even integers are closed under
multiplication, and likewise the odd integers. In particular, the square
of an even integer is even and the square of an odd integer is odd.
Now suppose that v2 were a rational number, say
vi
=
�,
b
where a and b are integers. We will presume, and this is essential for
the argument, that the rational fraction alb is in its lowest terms.
Specifically, we shall make use of the fact that a and b are not both even,
because if they were the fraction would not be in lowest terms. Squaring
the above equation, and simplifying, we get
2
a2
b2
= -,
a2
=
2b2.
The term 2b2 represents an even integer, so a2 is an even integer, and
hence a is an even integer, say a = 2e, where e is also an integer. Re­
placing a by 2e in the equation a2 = 2b2, we obtain
(2e)2
=
2b2,
4e2
=
2b2,
2e2
=
b2•
43
REAL N U M BERS
The term 2c2 represents an even integer, so b2 is an even integer, and
hence b is an even integer. But now we have concluded that both a and
b are even integers, whereas alb was presumed to be in lowest terms.
This contradiction leads us to conclude that it is not possible to express
v2 in the rational form alb, and therefore vi is irrational.
3.4 The Irrationality of v3
One of the proofs that v3 is irrational is similar to the proof of the
irrationality of v2 just given, except that here the key is divisibility by
3 rather than by 2. As a preliminary to the proof we establish that the
square of an integer is divisible by 3 if and only if the integer itself is
divisible by 3. To see this, we note that an integer divisible by 3 has the
form 3n, whereas an integer not divisible by 3 is of the form 3n + 1 or
the form 3n + 2. Then the equations
(3n)2
=
9n2
=
(3n + 1)2
=
9n2 + 6n + 1
(3n + 2)2
=
9n2 + 12n + 4
3(3n2),
=
3(3n2 + 2n) + 1,
=
3(3n 2 + 4n + 1) + 1
confirm the above assertion.
Next suppose that v3 were a rational number, say
v3
=
�,
b
where a and b are integers. Again, as in the v2 case, we presume that
alb is in lowest terms, so that not both a and b are divisible by 3.
Squaring and simplifying the equation, we obtain
3
a2 ,
b2
= -
a2
=
3b2•
The integer 3b2 is divisible by 3 ; that is, a 2 is divisible by 3 . So a itself
is divisible by 3, say a = 3c, where c is an integer. Replacing a by 3c in
the equation a2 = 3b2, we get
(3C)2
=
3b 2,
9c2
=
3b2,
3c2
=
b 2•
This shows that b 2 is divisible by 3, and hence b is divisible by 3. But
we have established that both a and b are divisible by 3, and this is
44
NU MBERS :
RATION A L
AND
IRRATION A L
contrary to the presumption that alb is in lowest terms. Therefore V3
is irrational.
3.5 Irrationality of v6" and
vi + VJ"
The proofs of the irrationality of vi and V 3 depended on divisibility
properties of integers by 2 and by 3, respectively, but the corresponding
proof for V6 can be made to depend on divisibility either by 2 or by
3. For example, if we parallel the vi proof, we would assume that
V6
=
'i '
where the integers a and b are not both even. Squaring, we would obtain
6
a2
b2
= - ,
a2
=
6b2.
Now, 6b 2 is even, so a 2 is even, so a is even, say a
write
a2
=
6b 2,
(2C)2
=
4C2
6b2,
=
6b2 ,
=
2e. Then we can
2C2
=
3b2 .
This tells us that 3b2 is even, so b2 is even, and thus b is even. But a
and b were presumed to be not both even, and so V6 is irrational. The
reader may, as an exercise, deduce the same conclusion by means of a
proof which is analogous to the V3 proof.
As a last example of irrationality in this chapter, we treat the case
v i + V3 by making it depend on the V6 case. Suppose that vi + V3
were rational, say r, so that
vi + V3 =
,. .
Squaring and simplifying, we get
2 + 2 V6 + 3
=
,. 2,
2 V6
=
,.2
-
5,
- ,. 2 - 5
V6 = -.
2
Now, rational numbers are closed under the four operations of addition,
subtraction, multiplication, and division (except by zero), and so
t(r2 - 5) is a rational number. But V6 is irrational, and thus we have a
contradiction. We can therefore conclude that vi + vl 3 is irrational.
REA L N U MBERS
45
Given any integer n = a · b of which we know that vn = va b
is irrational, we can deduce that the expression va + vb is irrational by
imitating the above proof.
.
Problem Set 10
I. Give two proofs that the square o f an integer i s d ivisible b y 5 if and only
if the integer itself is divisible by 5.
(a) First, give a proof parallel t o the analysis in the text in the case of
d ivisibility by 3 . Start from the fact that every integer is of one of the
five forms, 5n, 5n + I , 5n + 2, 5n + 3, or 5n + 4.
(b) Next, give a proof by use of the Fundamental Theorem of Arithmetic.
This theorem can be found in Chapter I or in Appendix B .
2. Prove that
3.
v5 is irrational.
Prove that v15 is irrational.
4. Prove that
V5 + V3 is irrational.
5. Prove that V2" is irrational.
6. Given that a (alpha) is an irrational number, prove that a-I
is irrational.
= l /a
also
7 Is 0 rational or irrational ?
3.6 The Words We Use
The language that we use to describe the various classes of numbers
is part of our historical inheritance, and so it is not likely to change
even though we may feel that some of the words are slightly peculiar.
For example, in everyday speech when we describe something as
"irrational," we usually mean that it is detached from good sense and
therefore unreasonable. But of course we do not regard irrational
numbers as being unreasonable. Apparently, the Greeks were surprised
when they discovered irrational numbers, because they had felt that,
given any two straight-line segments such as the side and the diagonal
of a square, there would be some integers a and b so that the ratio of
the lengths of the segments would be a/b. Thus the word "rational"
in its mathematical sense has reference to this ratio of whole numbers
and "irrational" refers to the absence of any such ratio.
The word "commensurable" also has been used to describe two
lengths whose ratio is a rational number. Two commensurable lengths
are so related that one can be "measured" by means of the other in
the following sense : If there exists some integer k such that when the
first segment is divided into k equal parts, each part of length I, it turns
46
N U M BERS :
I
2
I
3
I
2
I
3
RATIONAL
AND
I R RAT I O N A L
.... 1 ...
I
I
k-l k
0 0 0
....1 ....
o
0
I
m-i
0
Figure 1 1
m
out that the second segment is measured by a whole number, say m , of
such parts of length I, then the ratio of the lengths of the two segments is
k
kl
=
ml 111 '
i.e., rational (see Fig. 1 1 ). However, if the segments are such that the
ratio of their lengths is irrational (e.g. the side and the diagonal of a
sq uare), then the above construction can never be performed, no matter
how large we take k (and how small we take I) ! I n this case the given
segments are said to be incommensurable.
N umbers like V2, -0"24, or, in general, of the form {Y(i, where a is
rational and n is an integer, are called radicals.
The term "real numbers" is another inheritance from the past. If we
were to name them today, we would perhaps call them "one-dimensional
numbers." In any event, we do not regard numbers beyond the range
of real numbers as being "unreal." The reader is probably familiar with
the complex numbers, of which the real numbers constitute a subclass.
A complex number is a number of the form a + bi, where a and b are
real and i satisfies the quadratic formula i2 =
I . This definition is
introduced merely to round out the discussion of classes of numbers.
The scope of this book is limited to real numbers and so we are not
concerned with the wider class of complex numbers.
-
3.7 An Application to Geometry
Most high school textbooks on geometry leave a gap in some proof
or proofs where irrational numbers arise. The gap occurs when a result
is proved in the rational case only, with the irrational case left un­
finished. This happens often with the following result.
THEOREM 3. 1 . If three parallel lines are cut by two transversals, with
intersection points A, B, C. A ', B', C, as sholl'n in Fig. 12, then
AB
A'B'
'
BC = B'C
REAL
N U M BERS
47
where, (or example, A B denotes the length o( the line segment (rom A
to B.
Figure 1 2
This theorem can be used t o prove the basic theorem o n similar
triangles : if the three angles of one triangle are equal, respectively, to
the three angles of another, then corresponding sides are proportional
Figure 1 3
(Fig. 1 3). This result, i n turn, is often used to prove Pythagoras' Theo­
rem, and so trigonometry and analytic geometry are built on the basis
of these theorems.
We shall now prove Theorem 3 . 1 for the case where A B/Be is
irrational. We shall take Theorem 3. 1 for granted in the case where
A Bf Be is rational, because this part of the theorem ordinarily is proved
in books on elementary geometry. Before proving Theorem 3 . 1 in the
case where A B/ Be is irrational, it will be helpful to establish the follow­
ing preliminary result.
THEOREM 3.2. If m and n are positive integers such that
AB
m
-<-,
n
Be
48
N U M BERS :
RATIONAL
AND
IRRATIONAL
then
PROOF. We begin with a construction. Divide the line segment Be
into n equal parts, say each part of length a, so that Be = na. Then
mark off m more of these pieces of lengths a along the line segment BA,
A
A'
o
0'
-f---------"<B 7------� B'
c'
c
Figure 14
terminating at the point D. We first establish that D lies between B
and A, as in Fig. 1 4. Since Be = na and DB = ma, we can write
DB
Be
=
ma
na
=
n'
m.
by assumption
AB ,
-<Be
n
m
and so
DB
AB
·
<
Be
Be
This last inequality implies that DB < A B, because both fractions have
the same denominator BC Thus, since DB is shorter than A B, it follows
that D lies inside the line segment A B.
Next, from all these points of division we draw lines parallel to AA ' ,
with the point D ' corresponding to D on the right-hand side as in
REAL N U M BERS
49
Fig. 14. Thus by Theorem 3. 1 for the rational case (which we took for
granted), B'C' is divided into n equal parts, and D' B' into m equal parts
of the same length, so that
D'B'
m
B'C' = n '
However, from Fig. 14 we note that D'B' < A'B', and so we conclude
that
A 'B'
n < 8'C"
m
COROLLARY TO THEOREM 3.2. If
m
AB
m
A'B' .
then
n > Be '
n > B' C'
This corollary is analogous to Theorem 3.2, and so a similar proof
holds.
Thus we have proved Theorem 3.2 and a corollary ; we shall now use
these to prove Theorem 3.1 in the irrational case. Let 13 denote the
irrational number which represents the ratio AB/Be. We use the
decimal representation of 13 as in Section 3.2.
To illustrate what we are about to do, let 13 have the value
'
7r = 3 . 14t'59 . . . for example. Then we can write
3
4
< 13 < '
.
1
(1)
31
32
< 13 <
10
10 '
315
314
13
1 00 < < 1 00 '
3141
WOO
< 13 <
3 1 42 . . , etc.
1000 '
.
The fractions on the left are obtained by taking the numbers 3, 3 . 1 ,
3. 14, 3.141, etc. from the decimal expansion. The fractions o n the right
are obtained by increasing these same numbers by 1, 0. 1, 0.01, 0.001, etc.
The chain of inequalities ( 1 ) is infinite in number ; we have written
only the first four. These inequalities characterize the particular value
of i3 we are discussing, namely 7r. That is to say, if a number (3 satisfies
all the inequalities (1), then that number equals 7r.
The inequalities ( I ) were written in connection with an illustrative
50
N U M BERS :
RATIONAL
AND
I R RATIONAL
example, where fJ had the value 11". Now we drop this example, but we
point out that whatever irrational value (3 has, its decimal representation
will provide us with a chain of inequalities
a-l < I� < 1 + a l '
1'
1
1
--
(2)
a4 < {3 < 1 + a4 , . . . , etc.
-1000
--
1000
which characterizes (3 uniquely, and in each inequality {3 is between two
rational numbers. The symbols at. a2 , a3, . . . denote integers.
Our plan is to let {3' denote the ratio A' B'/B'C' and to prove that iJ'
also satisfies the inequalities (2), j ust as {3 does. But these inequalities
characterize the number {3, and hence {3' is identified with {3, so that
{3 =
A'B' = {3, .
AB
BC = B' C'
All that remains is to show that {3' satisfies all the ineq ualities (2). To
do this we use Theorem 3.2. First we take any of the fractions aI/I,
a2/ 10, a3/ 1OO, etc., say a3/1 00, and interpret this as the rational number
mIn of Theorem 3.2. Then the hypothesis of Theorem 3.2,
m
A B,
- < 11
BC
becomes
and this is valid because of the inequalities (2). Then Theorem 3.2 tells
us that
m < A'B' .
Ii B'C"
that is,
REAL
51
N U M BERS
Thus we see that tJ' satisfies
al
l
<
{3' ,
etc.
By an analogous use of the corollary to Theorem 3.2, we get the
i neq ualities
{3'
<
I + at ,
I
(3'
<
1 + a2
'
10
(3'
<
1 + a3
'
100
�'
I
fJ
<
1 + a4
1000 '
etc.
'
'
Hence {3 satisfies the inequalities (2) j ust as (3 does. Thus {3 = {3 and
the proof of Theorem 3. 1 is complete.
3.8 A Summary
In this chapter, we have indicated that every real number can be
associated with exactly one point on the "real line." We have also seen
that every real number has exactly one representation as an infinite
decimal (provided we exclude an infinite succession of zeros, i.e.,
terminating decimals, in our representations). This representation by
infinite decimals was applied in Section 3.7 to a key theorem in ele­
mentary geometry. In addition we have demonstrated the irrationality
of certain numbers such as vi2, vi), vi2 + vi3, etc. Our methods,
however, were rather piecemeal, and we did not give any very general
procedure for determining whether or not a given number is rational.
In the next chapter, we shall study irrational numbers in a much
more systematic way. We shall derive a method by which a large class
of numbers can be classified as irrational.
C H A PT E R F O U R
Irrational Numbers
In the course of this chapter and the next, we shall learn that the
real n umbers can be classified not only into rational and irrational
numbers, but also i nto two other categories. One category contains the
so-called algebraic numbers, i.e., those numbers which are solutions of
algebraic equations with integer coefficients, and the other includes all
remaining n umbers and these are called transcendental numbers. This
distinction will become more meaningful in what follows. We mention
at once, however, that some algebraic n umbers are rational and some
are irrational, but all transcendental numbers are irrational.
The over-all purpose of this chapter is to devise a systematic method
for determining whether or not a given algebraic n umber is rational.
(Actually, we shall not treat the class of algebraic numbers in its greatest
generality, but we shall apply our method to many examples.) But before
we derive this method, we shall study some simple properties of irrational
numbers.
4.1 Closure Properties
I n contrast to the rational numbers which were shown to be closed
under addition, subtraction, multiplication, and division (except by
zero), the irrational numbers possess none of these properties. Before
showing this, we prove a theorem which will enable us to manufacture
infinitely many irrational numbers from one given irrational number.
THEOREM 4. 1 . Let a be any irrational number and r any rational number
except zero. Then the addition, subtraction, multiplication, and division of'
r and a yield irrational numbers. A lso - a and a- I are irrational.
PROOF. These results are readily established by means of indirect
proofs. Suppose, to begin with, that - a were rational, say - a = r' ,
52
I R RATION A L N U MBERS
53
where r' denotes the presumed rational number. Then we would have
a = , , and , is also a rational number. Thus we have a contradic­
tion because a is irrational.
The theorem asserts that - a, a- I = l/a, a + ', a - ', , - a, 'a,
ai"� and ,/a are irrational. We have already treated - a. In order to
prove the irrationality of a- I , we observe that it is a special case of ,/a
with , = 1 . Thus there is no need to treat this case separately.
Let us prove the remaining six cases all at once, in wholesale fashion.
If one or more of these expressions were rational, then one or more of
the following equations,
_
'
_
'
a - , = '2,
a + , = '\.
,-a=
'3,
'a = '4,
a
r
=
's,
,
= ,
a '6
would hold, where ,\. '2, '3, '4, 'S. '6 denote some rational numbers.
Solving these equations for a, we get
a = '2 + ',
a = '1 - r,
a=,
- '3,
'4
a = -,
,
a = -·
'6
r
The right-hand members of these equations are rational n umbers be­
cause of the closure properties of rational numbers. But none of these
equations holds, since a is irrational. Hence it is not possible for any
of the numbers a + r, a - ', etc. to be rational. Thus the proof of
the theorem is complete.
By means of Theorem 4. 1 , we can construct a large class of irrational
numbers from one single irrational number, for example, from v2.
Applying each statement of the theorem, we can assert, for example, that
- v2 ,
1
� ,
v2
v2 + 5,
3
-
v2 ,
- 2 v2 ,
v2
T'
4
v2
are all irrational. Since there are infinitely many rational numbers which
we can use in each of the assertions of the theorem, it is clear that in­
finitely many irrational numbers can thus be manufactured.
Moreover, any one of the numbers so constructed, say for example
V2 + 5, can now be used as a new irrational number a in the theorem.
Therefore, infinitely many more irrational numbers, e.g.,
- v2 - 5,
V2 + 5 '
v2 + 8,
5v2 + 25,
v2 + 5
7
etc.
can be generated from that one number.
Are the irrational numbers closed under addition ? No, they are not.
N U M BERS :
54
RATION A L A ND IRRAT[0N A L
To prove this we need only exhibit two irrational numbers whose
sum is rational. Now, in the preceding chapter v2 was shown to be
irrational, therefore - v2 is irrational by Theorem 4. 1 . But the sum of
v2 and - v2 is 0, which is rational ; so is the sum of 3 + v2 and
5 - v2, for example. More generally, the sum of ,) + a and '2 - a
(where ') and '2 are rational and a is irrational) is rational.
The proposition that the irrational n umbers are not closed under
addition does not mean that if we add any two irrational n umbers the
sum will be rational. It means only that there is at least one case where
the sum is rational. The result obtained when two irrational numbers
are added may be either rational or irrational, depending on the two
n umbers we start with. Whereas the sum of v2 and - v2 is a rational
n umber, the sum of v2 and v3 is an irrational number, as we estab­
lished in the preceding ch'lpter.
Are the irrational numbers closed under subtraction? No, because,
for example, if we subtract v2 from itself we get the rational number O.
Similarly, the irrational numbers are not closed under mUltiplication
or division. These results are so similar to the earlier ones that we
leave the proofs to the reader in the next set of problems.
Problem Set 1 1
(In some of these problems, it may be helpful to use some of the results
derived in the preceding chapter, namely that v2, V3, V6, and V2 + v3
are irrational.)
I . Exhibit two irrational numbers whose difference is irrational.
2. Exhibit two irrational numbers whose product is rational and, so prove
that the irrational numbers are not closed under multiplication.
3. Exhibit two irrational numbers whose product is irrational.
4. Exhibit two irrational numbers whose quotient is rational, and so prove
that irrational numbers are not closed under division.
5. Exhibit two irrational numbers whose quotient is irrational.
* 6. Prove that
7.
Let
a
V3 (V6 -
3) is irrational.
be a positive irrational number. Prove that v� is irrational.
8. Given that a and {3 are irrational, but
and a + 2{3 are irrational.
a
+ {3 is rational, prove that
a -
{3
4.2 Polynomial Equations
It was proved in the preceding chapter that v2, v3, and v6 are
irrational. As might be expected (or, perhaps, as the reader already
I R RATION A L N U M BERS
55
knows) such numbers as "';7, �5, and �91 are also irrational. What
we want to do next is establish the irrationality of all such numbers by
a common scheme instead of treating them one at a time. To do this,
we shall shift the emphasis from the numbers themselves to simple
algebraic equations having the numbers as roots. For example, "';2 is
a root of the equation
x2 - 2 = 0; other ways of saying this are :
" "';2 is a solution of x2 - 2 = 0" or ""';2 satisfies the equation
x2 - 2 = 0." Similarly, the other n umbers under consideration satisfy
equations as follows :
"';3,
"';6,
"';7,
�5,
�91,
x2 - 3 = 0,
x2 - 6
x2 - 7
=
0,
=
0,
x3 - 5
=
0,
xS
=
o.
-
91
What we shall do is to establish that these equations, and more
generally all equations satisfying certain conditions, have no rational
roots. To begin we must define a few terms used to describe equations.
By a quadratic polynomial in x, we mean an expression of the form
ax2 + bx + c, where a, b, and c are called the coefficients. A cubic
polynomial, or a polynomial of degree 3, is of the form ax 3 + bx2 +
CX + d. In order to avoid introducing new letters as the degree is in­
creased, it is convenient to write
Now, a polynomial of any degree n (where n is a positive integer) has
the form
cnxn + Cn - lXn- 1 + . . . + CIX +
co,
with CII not zero.
A polynomial equation is a statement of equality of the form
C"x" + C,, _ IXn- 1 + . . . + CIX + Co
(I)
co, CJ , C2 ,
•
•
•
=
0;
, c" are called its coefficients.
EXAMPLE. Identify the values of n, en, etc. , when the equation
3x6 + 2x5 - x4 + I Ox3 + 4x - 7 = 0
is thought of as having the form ( I ) above.
56
N U M B ERS : RAT[0NA L A N D [ R RAT[ 0N A L
SOLUTION.
By direct comparison we see that
n = 6, c6 = 3,
cs = 2, c4 = - I ,
c3 = I O, C2 = 0, c ( = 4, co = - 7.
Observe that the requirement that the coefficients of eq. (1) be
integers is no stricter than the requirement that the coefficients be
rational ; for, if they are rational then Co = ao/bo, CI = a I !bJ , C2 =
a2/b2, . . . , where the a' s and b's are integers. All these fractions can be
written so as to have a common denominator, for example the product
bob 1 b2 · . . bn, by which we may multiply both sides of the equation
and thus obtain a new equation whose coefficients are integers and
whose roots are the same as those of the original equation.
We recall that a root of an equation in x is a value which, when
substituted for x, satisfies the equation. For example, as we noted
earlier, v7 is a root of x2 - 7 = O.
-
EXAMPLE.
Is 2/5 a root of IOx3 + 6x2 +
SOLUTION.
By substitution of 2/5 for x, we get
10
X
Gr + Gr + �
6
2 = O?
- 2 = 0,
and this is a correct statement of arithmetic. Therefore 2/5 is a root of the
equation.
We are now ready to return to the main point. We repeat that the
method we are about to develop for deciding whether or not a given
number is irrational can be applied if and only if we can write down a
polynomial equation which has the number under consideration as a
root. The method can be used not only for the numbers whose irration­
ality we established in the preceding chapter, but also for any number
which can be written as a finite combination of the symbols +, - , X ,
+ , and radicals �r of rational numbers. For example,
viv2 + v3 + y!\i7 - v7
1�25
is a complicated case of the kind of number we are talking about.
We do not prove in this book that all such numbers are roots of
polynomial equations with integer coefficients, but we shall write the
polynomial equations satisfied by many such numbers.
Problem Set 1 2
I . Identify the values of n , Cn, etc. , when the following equations are though t
of as having the form of eq. ( I ) above :
IRRATION A L N U MB ERS
(a) 1 5x3 - 23x2 + 9x - I = 0 ;
(b) 3x3 + 2x2 - 3 x - 2 = 0 ;
(c) 2x3 + 7x2 - 3x - 1 8 = 0 ;
(d) 2x4 - x2 - 3x + 5 = 0 ;
(e) 3x5 - 5x3 + 6x2 - 12x + 8
(0 x4 - 3x2 - 5x + 9 = O.
=
57
0;
2. (a) Is 1/3 a root of (a) above?
(b) Is -2/3 a root of (b) above?
(c) Is 3/2 a root of (c) above?
(d) Is 2 a root of (d) above?
(e) Is -2 a root of (e) above?
(0 Is 1/2 a root of (0 above?
3. Establish that V7 is a root of
� x2 - � = O.
4. Prove that if a number is a root of a polynomial equation such as
a 3 x3 + a2 x2 + a l x + ao = 0
b3
bl
b2
bo
with rational coefficients a 3/b 3, etc., then that number is a root of a poly­
nomial equation with integer coefficients.
5. Generalize the result of the preceding problem from equations of degree 3
to equations of degree n.
4.3 Rational Roots of Polynomial Equations
Our purpose now is to derive a simple rule, given as Theorem 4.3
below, which enables us to find all rational roots of any given poly­
nomial equation with integer coefficients. Th us we will be able to
separate the rational roots and the irrational roots of an equation, and
so establish the irrationality of a wide class of numbers.
But first we need the following auxiliary result.
THEOREM 4.2. Let u, v, w be integers such that u is a divisor of vlV, and
u and v have no prime factor in common. Then u is a divisor of w. More
generally, if u is a divisor of v'/w, where n is any positive integer and u
and v have no prime factor in common, then u is a divisor of w.
Before presenting the proof, let us illustrate this proposition with
some examples.
(1) Let u = 2, v = 3, and vw = 12. Now, 2 and 3 have no prime
factor in common. Also 2 is a divisor of 1 2, so the hypothesis of Theo­
rem 4.2 is satisfied. The conclusion that 2 is a divisor of w = 1 2/v = 4
is valid.
(2) Let u = 4, v = 5, v 3 w = 500. The numbers 4 and 5 have no
prime factor in common and 4 divides 500. The conclusion of the more
general statement, namely, that 4 divides w = 500/1 25 = 4, also holds.
N U M BERS : RATION A L A N 0 [ RRAT[ 0N AL
58
PROOF. The main ingredient in this proof is the Fundamental
Theorem of Arithmetic which is proved in Appendix B at the end of
the book and which assures us that there is only one way to factor
u, v, and w into prime factors. Since u divides vw, all the prime factors
of u occur also in vw .. moreover, if any prime p occurs in u to the power
a, then it occurs in vw to at least that same power, i.e., it occurs in VII'
to a power {3, where {3 :2: a. Now, since u and V have no prime factor
in common, it follows that all prime factors of u occur, to at least the
same power, in the factoring of w. Hence u is a divisor of w.
The last statement in the theorem can be argued in a similar way.
The assumption that u and V have no common prime factor assures us
that u and v' have no prime factor in common. Once again it follows
that v' in no way contributes to the fact that u is a divisor of v" w, and
so u must be a divisor of w.
We now have enough background material to state and prove the
following proposition.
THEOREM 4.3. Consider any polynomial equation with integer coefficients,
(1)
c"x" + Cn_ 1 X"- 1 + C,,_2X"- 2 + . . . + C2x 2 + CtX + Co
=
O.
If this equation has a rational root alb, where alb is presumed to be in
lowest terms, then a is a divisor of Co and b is a divisor of c" .
Again we shall illustrate this statement by an example before we
present its proof. Consider the equation
2x3 - 9x2 + lOx - 3
=
O.
The theorem says that, if alb is a rational root in lowest terms, then a is
a divisor of - 3 and b is a divisor of 2. Hence, the possible values for a
are + 1 , - 1, + 3 -3, and those of b are + 1 , - 1 , + 2, -2. Combining
these possibilities, we find that the following set contains all possible
roots :
,
+1
'
+1
+1
=1'
+1
'
+2
-1
'
+1
+1
-2'
+3
+ 1'
-1
'
-1
- 1'
+3
-1
+2'
+3
'
+2
-1
- 2'
+3
-2'
-3
+ 1'
-3
- 1'
-3
'
+2
-3
-i
This list contains only eight distinct numbers, namely, 1 , - I , 1/2,
- 1/2, 3, 3, 3/2, 3/2 Of these, only the n umbers 1, 1/2 , and 3 are
actual roots of the equation, as the reader may verify by substitution.
-
-
.
I RRATIONAL N U M BERS
59
PROOF. Let alb be a root of eq. (1). This means that if alb is sub­
stituted for x, then
is true. We begin by giving the proof for the special case where n = 3,
because it will be easier for the reader to follow. Then we shall give an
analogous argument for the general case.
In the case n = 3, eq. (2) is simply
M ultiplying by b 3 , we get
(3)
First we write eq. (3) in the form
or
This shows that b is a divisor of C3a 3 . We apply Theorem 4.2 at this
point, with u, v, and w replaced by b, a, and C3 , respectively. The
hypothesis of Theorem 4.2, that u and v have no prime factor in com­
mon, is satisfied because alb is in lowest terms so that a and b have no
prime factor in common. Hence we can conclude from Theorem 4.2
that b is a divisor of C3 . This is a part of the desired conclusion in
Theorem 4.3, because in this case n = 3, so that c" is C3 .
Next we write eq. (3) in the form
or
This shows that a is a divisor of cOb 3 • By an argument virtually identical
to the earlier one, i.e., by again applying Theorem 4.2, we conclude
that a is a divisor of co. Thus the proof is complete in case n = 3.
60
N U M BERS : RATIONAL AND I RRATIONAL
To prove the theorem for any n, we return to eq. (2) and multiply it
by b" to obtain
Now (4) can be rewritten as
or
This shows that b is a divisor of c"a". We apply Theorem 4.2 with u, v,
and w replaced by b, a, and cn, respectively, and conclude that b is a
divisor of CII
Next we rewrite eq. (4) as
•
This shows that a is a divisor of cob". Again by applying Theorem 4.2
with u, v, w replaced by a, b, co, respectively, we conclude that a is a
divisor of co. This completes the proof of Theorem 4.3.
We could have avoided the argument of the last paragraph by
observing that there is a symmetry about eq. (4), and that b stands in
relation to C II in the equation exactly as a stands in relation to Co.
Next we examine the situation that holds when c" = 1 .
COROLLARY 1 . Consider an equation of the form
whose coefficients are integers. If such an equation has a rational root,
i! is an integer; moreover, this integer root is a divisor of co.
PROOF. Consider any rational root a/b. We may presume that b is a
positive integer, because if b were negative we could absorb the minus
sign in a. According to Theorem 4.3, b must be a divisor of cll ; that is,
b must be a divisor of 1. But + 1 and - 1 are the only divisors of 1 ,
and s o we must have b = + 1 , since we ruled out any negative value
for b. Consequently any rational root is of the form a/I , and so it is an
integer a. Also by Theorem 4.3 we know that a is a divisor of co, and
thus the proof of the corollary is complete.
I RRATIONA L N U M BERS
61
Y7 is irrational.
Y7 is a root of x2 - 7 =
Prove that
EXAMPLE.
O. Here, according to our notation,
1 and Co = - 7.
Now, there are two ways we can proceed. One way is to use Corollary 1
and say : If x2 - 7 = 0 has a rational root alb, then that rational root would
have to be an integer. We can show that Y7 is not an integer and that therefore
it is not a rational root of x2 - 7 = O. Hence it must be an irrational root.
Clearly, V7 is not an integer because it lies between the consecutive integers
2 and 3 ; this, in turn, follows from the inequalities
SOLUTION.
C2
=
4<
7 <
9,
2 < v7 <
3.
V4 < V7 < V9,
Another way employs Corollary 1 in its full form, according to which any
rational root of x2 - 7 = 0 is an integer which is an exact divisor of - 7.
The only divisors of -7 are I , - I , 7, and - 7. But none of these is a root
as can be seen by simple verification; the equations
] 2 - 7 = 0,
72 - 7 = 0,
( - 7)2 - 7 = O.
( _ 1)2 - 7 = 0,
are all false. Hence x2 - 7 = 0 has no integral root, hence no rational root,
and V7 is an irrational number.
EXAMPLE.
Prove that
-\Is is irrational.
SoLUTION. -\Is is a root of x3 - 5 = O. According to Corollary 1 , if this
equation has a rational root it is an integer which is a divisor of 5. The divisors
of 5 are + 1 , - I , +5, and - 5 . But none of these is a root because the
equations
]3 - 5
=
0,
( _ 1)3 - 5
are all false. Hence x3 - 5
=
=
0,
53 - 5
=
0,
( - 5)3 - 5
0 has no rational roots, and so
0
-\Is is irrational.
=
These two examples are special cases of the following more general
result :
COROLLARY 2 . A number of the form �a, where a and n are positive
integers, is either irrational or it is an integer," in the latter case a is the
nth power of an integer.
PROOF. This follows from Corollary I because �a is a root of
a = 0, and if this equation has a rational root, it must be an
integer. Furthermore, if �a is an integer, say k, then a = kll .
X"
-
Problem Set 1 3
V2, Y3, yO, and {/91 are irrational.
2. Prove that (4YO
3)/6 is irrational.
I . Prove that
-
62
N U M BERS : RATION A L AN D I RRATl0N AL
3. Prove that vB is irrational.
3v'15) is irrational.
4. Prove that 4/( 16
5. Prove that v6 is irrational.
-
6.
Prove that ( I /3)(2{/6 + 7) is irrational.
7. Prove that Theorem 4.3 becomes a false statement if the words "pre­
suming that alb is in lowest terms" are omitted.
4.4 Further Examples
In Chapter 3, it was proved by a method which applies to a rather
wide class of numbers that V2 + vj is irrational. However, an even
wider class of numbers can be treated with the aid of Corollary 1 .
Let us discuss V2 + V3 again. I f we write x = V2 + V3, then we
have
x - V2
=
V3.
Squaring both sides, we obtain
x2 - 2xV2 + 2
=
3,
and by rearranging terms, we have
x2 - 1
=
2xV2.
If we square this again, we obtain
X4 - 2x2 + 1
=
8x2
or
X4
(5)
-
IOx2 + 1
=
O.
Because of the way in which eq. (5) has been constructed, we know
that V2 + vj is a root. Next, we shall apply Corollary 1 to show that
eq. (5) has no rational roots, and from this we shall conclude that
V2 + vj is irrational.
The application of Corollary 1 to eq. (5) tells us that if this equation
has any rational roots, they must be integers which divide 1 . But the
only divisors of 1 are + 1 and - 1, neither of which is a root of
x4 - IOx2 + 1 = O. We thus conclude that eq. (5) has no rational
roots and that V2 + V3 is irrational.
Another way of reaching the same conclusion is this : Instead of test­
ing whether + 1 and 1 are roots of eq. (5), we may argue as follows.
-
I RRATIONAL N U M BERS
63
Even if + 1 or - 1 , or both, were roots of eq. (5), we can observe that
the root V2 + vj is different from + 1 and from - 1 ; for example,
we can argue that both vi and vj are larger than 1 , so that their sum
is too large to be equal to + 1 or - 1 . Hence V2 + vj is not among
the possible rational roots of eq. (5), regardless of whether or not + 1
or - 1 are actual roots. It follows that V2 + vj is irrational.
Prove that {l2 - V3 is irrational.
SoLUTION. Writing x = v2 - v3, we see that
x + v3 = v2.
Now, cube both sides and obtain
x3 + 3V3x2 + 9x + 3 v3
When the terms are rearranged,
EXAMPLE.
=
2.
x3 + 9x - 2 = - 3 V3(x2 + I ).
By squaring, we get
xl> + 18x4 - 4x3 + 81x2 - 36x + 4 = 27 (x4 + 2x2 + I )
or
xl> - 9x4 - 4x3 + 27x2 - 36x - 23 = O .
This equation has been so constructed that v2 - v3 is a root. But the
only possible rational roots of this equation are integers which are divisors
of - 23. Hence the only possible rational roots are + 1 , - I , +23, and 23,
and these are not roots, as straightforward substitution shows :
(False!)
+ I : 16 - 9( 1 )4 - 4( 1) 3 + 23(1)2 - 36( 1 ) - 23 = 0
- I : ( - 1)6 - 9( _ 1 )4 - 4( _ 1)3 + 27( _ 1 )2 - 36( - I ) 23 = 0 (False!)
(23)6 - 9(23)4 - 4(23) 3 + 27(23)2 - 36(23) - 23 = 0
23 :
(False, because for example, (23)6 is much too large to be
canceled out by the other terms !)
-23 : ( - 23)6 - 9( -23)4 - 4( -23) 3 + 27( -23)2 - 36( - 23) - 23 = 0 (False!)
Hence there are no rational roots and so {l2 - v'3 is irrational.
As in the preceding example, there is no need to test whether + 1 , - 1 , + 23,
and - 23 are roots of the equation. Instead, we can argue that �i vj is
different from any of these four possible rational roots. We observe that �2
is in the vicinity of 1 .2 and Vj" is in the vicinity of 1 .7. Consequently �i vj
is approximately -0.5, and hence not equal to any of the values + 1 , - I ,
+23, or -23. It follows that the root � i Vj" is irrational, since i t is
different from all the possible rational roots.
-
-
-
-
-
Problem Set 1 4
I . Prove that Vj" - vi i s irrational.
2. Prove that {/j" + vi is irrational.
3. Prove that {Is
Y3 is irrational.
-
64
N U M BERS : RATIO N A L A N D I RRATIO N A L
4.5 A Summary
In this chapter we have been dealing with so-called "algebraic
irrationalities." We have seen that there are infinitely many irrational
numbers and we have studied ways of constructing some of them from
a given irrational number.
We have also found the following method for testing whether or not
a given number k is irrational :
First we look for a polynomial equation
satisfied by the value x = k. (If we cannot find such an equation,
we cannot apply this method.)
Next we apply Theorem 4.3, or, if en = 1 , Corollary 1 . Often it
is clear that the equation possesses no rational roots at all. Then k
is clearly an irrational root. Sometimes, we see at a glance that k
is different from all the possible candidates for rational roots of
the equation, and so we may deduce the irrationality of k. Or, by
direct substitution, we select from all the candidates those rational
numbers which are actual roots of the equation. Then, in order to
prove that k is irrational, we must show that k is different from
all the rational roots.
In the next chapter we shall use the methods of this chapter to
demonstrate the irrationality of many trigonometric numbers, and we
shall use the Fundamental Theorem of Arithmetic to demonstrate the
irrationality of many logarithmic numbers. Moreover, we shall learn
that there are irrational numbers which are not roots of algebraic
equations.
C H A P T E R F I V E
Trigonometric and Logarithmic
Numbers
The reader is undoubtedly familiart with such trigonometric functions
as sin () and cos (), and knows that each of these functions assigns a real
number to every angle (). He has probably also encountered the loga­
rithmic function log x which assigns a real number to every positive real
number x.
The values of the trigonometric functions are, except for certain
special values of the angle (), irrational ; t similarly, the values of log x
are irrational for almost all positive real numbers x.
We cannot prove these assertions in their full generality here, so we
shall confine our attention to some simple examples.
5.1 Irrational Values of Trigonometric Functions
We shall show, using the methods of the previous chapter and certain
basic trigonometric identities, that for many angles () the corresponding
values of the trigonometric functions are irrational.
To this end, let us first recall the following basic trigonometric
formulas :
(1)
cos (A + B)
=
cos A cos B - sin A sin B,
(2)
sin (A + B)
=
sin A cos B + cos A sin B.
t Readers who have not yet studied trigonometry or logarithms can find an intro­
duction to these subjects in Plane Trigonometry by A. L. Nelson and K. W. Folley,
Harper, 1956.
t If they are listed in decimal representation in a table, then the irrational numbers
which have infinite decimal expansions have been cut off at some point. In other
words, such a table gives approximations to the numbers.
65
N U M BERS : RATIONAL A N D I RRATIONAL
66
Replacing A and B by a single value, say 8, we get
cos 28 = cos28 - sin28,
(3)
sin 2() = 2 sin 8 cos 8.
(4)
Next, if we replace A by 28 and B by 8 in ( 1 ), we get
cos 38
=
cos 28 cos 8 - sin 28 sin 8
Using (3) and (4) and also the well-known identity cos28 + sin28
we obtain
cos 38
=
(cos28 - sin28) cos 8 - (2 sin 8 cos 8) sin 8,
=
cos38 - 3 sin28 cos 8,
=
cos38 - 3(1 - cos28) cos 8
=
1,
or
(5)
cos 38
=
4 cos38
-
3 cos 8.
Now consider the number cos 20°. By setting 8
=
20° in (5), we have
I f we write x for cos 20°, and make use of the fact that cos 60°
we get
1
2
-
=
=
t,
4x3 - 3x
or
(6)
8x 3 - 6x - 1
=
o.
Because of the construction of eq. (6), we know that cos 20° is a root.
Applying Theorem 4.3 to eq. (6), we see that the only possible rational
roots of this equation are ± 1 , ± t, ± t, ± t. But none of these eight
possibilities is an actual root, as can be seen by substitution into eq. (6).
Hence we conclude that eq. (6) has no rational roots and so cos 20° is
an irrational number.
This conclusion can also be reached without testing whether the
T R I G 0 N O M ET R I C N U M BE R S A N D L O G S
67
possible rational roots ± I , ± t, ± t, ± i are actual roots of eq. (6).
It is enough to show that cos 20° is different from all these eight values.
This we can do by looking at the value given for cos 20° in a table of
trigonometric functions. (Such a table gives an approxi mati on only,
of course.) Or, we can observe that cos 20° lies between cos 0° and
cos 30°, the cosine being a decreasing function for these angles. Thus we
see that cos 20° lies between 1 and v3/2 ; thus it is between 1 and 0.8.
It follows that cos 20° cannot be equal to an y of the possible rational
roots of eq. (6), and therefore cos 20° is an irrational number .
EXAMPLE.
Prove that sin 1 0° is irrational.
FIRST SOLUTION. One way to solve this problem would be to begin with the
trigonometric identity for sin 39, i.e.,
sin 39 = 3 sin 9 - 4 sin39,
(7)
which can be obtained from (2) in the same way that (5) was obtained from ( 1 ) .
Replacing 9 by 10° in (7), and using the fact that sin 30° = 1 /2, w e get
!2
- 3 sin 10° - 4 sin3 1 O°.
Write x for sin 1 0° to obtain
I
= 3x - 4x3'
2
-
8x3 - 6x + I
=
O.
As with eq. (6), it is not difficult to show (by Theorem 4.3) that the equation
8x3 - 6x + I = 0 has no rational roots. Hence sin 1 0° is irrational.
SECOND SOLUTION.
(8)
cos 29
Eq. (3) has two alternative forms.
=
2 cos29 - I ,
cos 29
= I
-
2 sin29,
both of which can be obtained from (3) by use of the basic identity
sin29 + cos29 = I .
If, i n the second alternative of (8), we replace 9 by 1 0°, we get
cos 20" = I - 2 sin2 1 O°.
(9)
Now suppose that sin 10° were rational. Then both sin2 1 0° and I - 2 sin2 1 O '
also would be rational. But cos 20° is irrational, as we have already proved.
Hence we have a contradiction, so sin 1 0° is irrational.
Problem Set 1 5
I n solving these problems, make use (wherever helpful) of any previous
results, either in the text or in the problems themselves.
I . Prove that the following numbers are irrational :
(a) cos 40°
2. Establish identity (7).
(b) sin 20°
(c) cos 10°
(d) sin 50°
68
N U M BERS : R ATIONA L A N D I RRATIONA L
(a) Establish the identity cos 5(]
1 6 cos5(]
(b) Prove that cos 12° is irrational.
4. Which of the following are rational?
3.
=
(a) sin 0°
(b) cos 0°
(c) tan 0°
(d) sin 30°
(e) cos 30°
(f) tan 30°
-
(g) sin 45°
(h) cos 45°
(i) tan 45°
20 cos3(] + 5 cos (].
sin 60°
(k) cos 60°
(I) tan 60°
(j)
5.2 A Chain Device
The methods used in Section 5. 1 can be extended to prove the
irrationality of the trigonometric functions of any angle which is a
whole number of degrees, minutes, and seconds, with but a few obvious
exceptions. Thus we are speaking of angles like 14°4 1 '33". Exceptions
must be made for the angles 0°, 30°, 45°, 60°, and also any angles
obtained from these four by adding or subtracting any integral multiple
of 90°. This is not to say that all the trigonometric functions of 30°, for
example, are rational, but at least one trigonometric function of 30°
is rational.
These assertions will not be established in their broadest generality,
because the equations arising for such numbers as cos 1 4°4 1 ' 1 3" are
too complex to be brought within our purview. Nevertheless there is
one simple principle that will carry us a long way, namely :
Jf(] is any angle such that cos 2(] is irrational, then cos (], sin (], and tan (]
are also irrational.
To prove this, we first use eq. (8). Suppose that cos (] were rational.
Then cos2(] and 2 cos2(]
1 also would be rational. But 2 cos2(]
1 is
cos 2(], which is irrational.
Similarly, suppose that sin (] were rational. Then sin2(] would be
rational and so would 1
2 sin 2(]. But this is cos 2(] again.
Finally, suppose that tan (] were rational. Then tan2(] would be
rational, and then we could use the well-known trigonometric identity
-
-
-
1 + tan2(]
=
sec2(]
1
cos2(]
= -­
to see that cos2(] would be rational. But again it would follow from
eq. (8) that cos 2(] is rational, and so we have a contradiction. Hence
tan (] must be irrational.
By repeated application of the principle we have just proved, we
can establish the irrationality of infinitely many trigonometric numbers.
69
T R I G 0N O M ET R I C N U M BE RS A N D LOG S
For example. from the irrationality of cos 20°, we conclude that
cos
cos
cos
cos
cos
10°
5°
2°30'
lOIS'
37'30"
sin
sin
sin
sin
10°
5°
2°30'
1 ° 1 S'
Sin
37'30"
tan
tan
tan
tan
tan
10°
5°
2°30'
1 ° 1 5'
37'30"
are irrational.
Problem Set 1 6
I.
Prove that the following numbers are irrational
tan 1 5'"
sin 1 5 ,
(a) cos 1 5 ',
(b) cos r30',
tan r30',
sin 7"30',
(c) cos 22°30',
tan 22°30',
sin 22°30',
*(d) cos 35' ,
*(e) cos 25' ,
sin 35°,
sin 25',
tan 3 5 ',
tan 25".
2. Prove that 14"'41 ' 1 3" equals some rational number multiplied by 90-, i .e.
prove that 1 4°4 1 ' 1 3" is a rational multiple of 90 ' .
3. (a) Prove that if cos () is rational, then cos 3 () i s also rational .
(b) ]s this equivalent to proving that if cos 3() is irrational, then
irrational ?
4.
,
cos () is
Prove that i f sin 3() i s irrational, then sin () i s irrational.
5.3 Irrational Values of Common Logarithms
All the logarithms discussed in this book will be taken to the base
10, so there will be no need to specify this base in each case. We recall
that, given a positive real number y, its logarithm to base 10 is defined
to be a number k such that 1()A = y. Thus for any y > 0,
log y
=
k
and
10"
= y
are equivalent statements. All the proofs will be based on the Funda­
mental Theorem of Arithmetic, proved in Appendix B, that any integer
has a unique factorization into primes.
EXAMPLE I.
Prove that log 2 is irrational.
70
N U M BERS : RATIONAL AND I RRATIONAL
SOLUTION. Suppose, on the contrary, that log 2 = a/b . where a and b are
positive integers. It ·is reasonable to take a and h positive, because log 2 is
positive. This equation means that
2 = I OQ/b•
Raising both sides to power b, we get
2h = IOQ = 2Q5Q•
This is an equality between positive integers, so the Fundamental Theorem of
Arithmetic is applicable. In fact the Fundamental Theorem shows that this
equation cannot hold because 2b is an integer which is not divisible by 5 what­
ever value b may have, whereas 2Q5a is divisible by 5 since a is a positive
integer. Hence log 2 is irrational.
EXAMPLE
2. Prove that log 21 is irrational.
SoLUTION. Suppose, on the contrary, that there are positive integers a and b
such that
log 21
=b
a
or
Again we raise both sides to power b to get
21b = IOQ•
But this statement of equality cannot be true since 21b has prime factors 3
and 7, whereas 10" has prime factors 2 and 5.
EXAMPLE 3. Let e and d be two different non-negative integers. Prove that
log (2C5� is irrational.
SOLUTION. Again we use an indirect argument. Because of the conditions
imposed on e and d, we know that 2c5d exceeds I , so that log (2cSd) is positive.
Suppose that
log (2C5d) =
�,
where a and b are positive integers. Then we know that
2c5d = IOQ/b•
Raising both sides to power b, we get
2bc5bd = IOQ = 2Q5Q•
According to the Fundamental Theorem of Arithmetic, this equation holds
only if be = a and bd = a, i.e., if be = bd. But since e and d are different
integers, so also are be and bd. Thus log (2C5d) is irrational.
Problem Set 1 7
1 . Prove that log 3/2 is irrational.
2. Prove that log 1 5 is irrational.
3. Prove that log 5 + log 3 is irrational.
* 4. Prove that the integers 1 , 2, 3, . . . , 1000 can be divided into three distinct
non-overlapping classes.
T R I G O N O M ET R I C N U M B E R S A N D L O G S
71
Class A : the integers I , 10, 100, 1000 ;
Class B: integers of the form 2c5d, where c and d are unequal ;
Class C: integers divisible by at least one odd prime p, with p not equal
to 5 ;
and that log n is rational if and only if n is in Class A .
5.4 Transcendental Numbers
Besides the classification of real numbers into rational and irrational,
there is another separation into algebraic and transcendental. If a real
number satisfies some equation of the the form
with integral coefficients, we say that it is an algebraic number. If a real
number satisfies no such equation, it is called a transcendental number.
(Complex numbers are divided into algt'braic and transcendental
numbers in exactly the same way, but we will confine our attention to
real numbers.)
It is easy to see that every rational number is an algebraic number.
For example, 5/7 satisfies the equation 7x
5 = 0, and this equation
is of the prescribed type. M ore generally, any rational number alb
satisfies the equation bx - a = 0, and so it is an algebraic number.
Since every rational number is algebraic, it follows that every non­
algebraic number is non-rational (see [12] of "Twelve Ways of Stating
'If A then B ' , p. 27), or, stated more conventionally, every transcen­
dental number is irrational. We can illustrate this schematically as in
Fig. ] 5.
In the figure, we have listed v2 and ..:;:; as examples of algebraic
numbers. They are algebraic because they satisfy the algebraic equations
-
"
x2
-
2
=
0
and
x3 - 7
=
0
respectively. On the other hand, the numbers log 2 and 7r have been
listed as transcendental numbers. (The number 7r, with value 3. 14159 . . .
is the ratio of the length of the circumference to the length of the
diameter of any circle.) We cannot prove here that these are trans­
cendental numbers because such proofs involve much deeper methods
than we are using. The transcendence of 7r has been known since 1 882,
but the transcendence of 2V2 and log 2 are much more recent results,
known only since 1934. The number 2vi was used as a specific example
by the great mathematician David Hilbert when, in 1900, he presented
a famous list of twenty-three problems which he viewed as the outstand-
72
N U M BERS : RATIONA L A N D I RRATION A L
Rational (All these are algebraic numbers.)
Real numbers
/
�
Irrational
Algebraic, e.g., Vi and .,y-;
(
Transcendental, e.g., 2v2, log 2, and
7r
<..
Rational
/
�
Algebraic
Real numbers
Irrational
Transcendental (All these are irrational numbers.)
Figure 1 5
ing unsolved mathematical questions. Specifically, Hilbert's seventh
problem was to decide whether aR is algebraic or transcendental, given
that a and {3 are algebraic numbers. (The cases a = 0, a = 1 , and (3
rational were excl uded because in these cases it is rather easy to prove
that afJ is algebraic.) I n 1934, it was settled by A. Gelfond and inde­
pendently by Th. Schneider that aR is transcendental. Of course, the
transcendence of 2vi is a special case of the general result. Another
special consequence is the transcendence of log 2 ; for, if we denote
log 2 by {3, and 10 by a, then by the definition of common logarithm,
If {3 were algebraic and irrational , then by the Gelfond-Schneider
Theorem, 2 would be transcendental. But this is not the case, so {3 = log 2
is either rational or transcendental. We have seen that it is not rational.
Hence log 2 is transcendental.
More generally, the Gelfond-Schneider theorem establishes the trans­
cendence of log r, provided that r is rational and log r is irrational. I n
view o f what we j ust proved in Section 5 . 3 (see also Problem 4 o f Set
1 7), this says that log r is transcendental when ,. is any positive rational
number except the following :
TRIG0N O M ETR I C N U M BE RS A N D LOGS
73
It should be kept in mind that all logarithms mentioned in this book are
common logarithms, that is logarithms to base 10.
Thus the numbers log n are transcendental if n is any integer between
I and WOO except n = 1 , n = 10, n = 100, and n = WOO On the other
hand, the trigonometric numbers like cos 20° that were proved irra­
tional in the early part of this chapter are algebraic numbers. The
general result is this : let r be any rational number and let (90rt denote
the angle obtained by multiplying 90° by r. Then
.
sin (90rt,
cos (90rt,
and
tan (90rt
are algebraic numbers. (The only qualification that must be put on
this statement is in the case of tan (90rt, where the rational number r
must be restricted to values for which this trigonometric function exists
as a real number. For example, r = 1 is not admitted because tan 90°
is not a real n umber.)
We said above that 'If" is a transcendental number. This implies that
'If" is an irrational number, and although it is easier to prove that 'If" is
irrational than that it is transcendental, even this is beyond the scope
of this book.
Problem Set 1 8
I.
'" 2.
Prove that the following numbers are algebraic :
(a) V),
(b)
Vs,
(c) V2 + V),
(d) cos 20°,
(e)
sin 10°
Assuming that 'If" is a transcendental number, prove that 2'1f" i s a transcen­
dental number.
5.5 Three Famous Construction Problems
The theory of algebraic and transcendental numbers has enabled
mathematicians to settle three well-known geometric problems that
have come down from antiquity. These three problems, commonly
referred to as "duplicating the cube," "trisecting an angle," and
"squaring the circle," consist in performing the following constructions
by means only of the straightedge and compass methods of Euclidean
geometry :
( 1 ) "Duplicating the cube," or "doubling the cube," means to con­
struct a cube having twice the volume of a given cube. Although a
cube is a figure in solid geometry, the problem is really one in plane
geometry. For, if we take an edge of the given cube as the unit of
length (Fig. 1 6), the problem is to construct a line of length Vi2, because
74
N U M BE R S . R A T I O N A L A N D I R R A T I O N A L
1
1
Figure 1 6
this would be the length of the edge of a cube that ha'i twice the volume
of the given cube.
(2) "Trisecting an angle" means to devise a method, using only the
prescribed tools, by which any angle can be trisected. There are certain
special angles, 45° and 90° for example, which can be trisected with
straightedge and compass ; but the so-called "general" angle cannot be
divided into three equal angles with the prescribed tools.
(3) "Squaring the circle" means to construct a square equal in area
to a given circle or, equivalently, to construct a circle equal in area to
a given square.
These three constructions are now known to be impossible; that is,
they cannot be done by the prescribed straightedge and compass meth­
ods of Euclidean geometry. M any amateurs continue to work on these
problems, not knowing that their efforts are in vain. Although these
amateurs are aware that no mathematician has yet been able to achieve
these constructions, they are apparently not aware that the construc­
tions have been proved impossible. What many an amateur mathe­
matician achieves from time to time is an approximate solution to one
of these problems, but never an exact solution. The distinction here is
clear : the duplication of the cube problem, for example, is to produce
a construction that would give, with theoretically perfect drawing
instruments, a line not almost of length ":;2, but exactly of length ":;2 .
The problem is not solved, for instance, by a construction giving a
line of length 10(8
V62), even though the numbers 10(8 v62)
and ":;2' agree to six decimal places.
A special source of misunderstanding occurs in the case of the angle
trisection problem. It is possible to trisect any angle if markings are
allowed on the straightedge. Thus the assertion of the impossibility of
the trisection of a general angle can only be made with the understand­
ing that the allowable construction processes involve the compass and
the unmarked straightedge.
Because of the considerable confusion surrounding these three classic
problems, we now give a rough notion of how it can be established that
-
-
T R I G 0N O M ET R I C N U M BE R S A N D LOG S
75
the constructions are impossible. Because the details become rather
technical, we cannot give complete proofs. Nevertheless we hope to
make the matter plausible. If any reader wants to tackle it, there is a
full treatment of the trisection of the angle and the doubling of the cube
in R. Courant and H. Robbins, What Is Mathematics ?, Oxford Univer­
sity Press, pp. 127-1 38. The proof of the impossibility of squaring the
circle is much more difficult than the other two proofs.
How is it possible to prove that these constructions are impossible?
The first step is to get some idea of what kinds of lengths can be con­
structed with straightedge and compass, given a unit length. We state
without proof (anyone familiar with geometric constructions will
realize that the assertion is reasonable) that among the lengths that can
be constructed are successions of square roots applied to rational
numbers, for instance,
These numbers are all algebraic. The four that are listed as examples
in (10) are roots respectively of the equations
( 1 1)
x2 - 2
(12)
x4 - 2x2 - 1
=
0,
=
0,
x8 - 20x6 + 1 32x4 - 320x2 + 94 = 0,
x l C! - 8x l 4 + 8x l 2 + 64x l O - 98x 8
- 1 84x6 + 200x4 + 224x2 - 1 1 3 = O.
( 1 3)
(14)
Let us choose one of these, say ( 1 3), and verify it. We begin with
x
=
\) 5 - n / 1 + v 2.
Squaring, we get
x2
Moving
a
=
5 - 3 v' 1 + v2.
term and squaring again, we obtain
x2 - 5
=
- 3 v' 1 + v 2,
x4 - lOx2 + 25
=
9 + 9 v2,
x4 - lOx2 + 1 6
=
9 v2.
76
N U M BERS :
R A TI O N A L A N D
IRRATIONA L
A final squaring of both sides leads to ( 1 3).
Now, not only are the numbers ( 10) roots of eqs. ( 1 1) to (14), but
none of these numbers is a root of an equation of lower degree with
integer coefficients. Take the number VI + v2, for example. It
satisfies eq. ( 1 2) of degree 4, but it satisfies no equation of degree 3, 2,
or I with integer coefficients. (We do not prove this assertion.) When­
ever an algebraic number is a root of an equation of degree n with
integer coefficients, but is a root of no equation of lower degree with
integer coefficients, we say that it is an algebraic number of degree n.
Thus the numbers (10) are algebraic numbers of degree 2, 4, 8, and 16
respectively. This suggests the following basic truth about lengths that
can be constructed by the methods of Euclidean geometry :
THEOREM ON GEOMETRIC CONSTRUCTIONS. Beginning with a line seg­
ment of unit length, any length that can be constructed by straightedge
and compass methods is an algebraic number of degree 1 , or 2, or 4, or 8,
. . . , i.e., in gen:?ral, an algebraic number whose degree is a power of 2.
If the reader will take this result for granted, we can indicate how it
happens that the three famous constructions are impossible. t
Let us begin with the duplication, or doubling, of the cube. As we
saw when we stated the problem, it amounts to constructing a line of
length V"2 from a given unit length. But is V"2 a constructible length?
It satisfies the equation
( 1 5)
x3
-
2 = 0,
and this suggests that V"2 is an algebraic number of degree 3. I ndeed
this is so, and in order to prove it we must establish only that V"2
satisfies no equation of degree 1 or degree 2 with integer coefficients.
Although this is not difficult, it is a little tricky, and we shall postpone
this proof until the next section.
Since V"2 is an algebraic number of degree 3, by the Theorem on
Geometric Constructions, stated above, it is not constructible. Hence
we conclude that it is impossible to duplicate the cube.
Consider next the problem of the trisection of the angle. To establish
that this is impossible, it is enough to show that a specific angle cannot
be trisected by the prescribed methods. The specific angle that we take
is 60°. To trisect an angle of 60° means the construction of a 20° angle.
t The reader will recall (see p. 27, ( 1 2)) that this theorem implies the statement :
A lgebraic num bers of degree "', where III is not a power of 2, are not constructible
by straightedge and compass ; also t ranscendental numbers are not so constructible.
T R [ G 0N O M ETR [ C N U M B ER S A N D L O G S
77
This amounts to constructing, from a given line segment of length 1 ,
a line segment whose length i s cos 20°. To see this, consider a triangle
of base 1 with base angles 60° and 90°, as shown in Fig. 17. Thus we
c
Figure 1 7
have a triangle A BC with base A B = 1 , angle BA C = 60°, angle
= 90°. On the line BC, let the point D be chosen in such a way
that angle BA D = 20°. From elementary trigonometry, we know that
A BC
AD
=
AD
-1
=
AD
AB
=
sec 20° .
Thus the trisection of the 60° angle amounts to the construction of a
line segment of length sec 20°. But this in turn amounts to the construc­
tion of a line segment of length cos 20°, because cos 20° and sec 20° are
reciprocals, and it is well known that if a certain segment is constructible,
then the segment of reciprocal length is also constructible.
So the question is : Can a line segment of length cos 20° be con­
structed from a given line segment of length I ? We know from eq. (6)
that cos 20° is a root of a cubic equation, i.e., an equation of degree 3.
Moreover we state (without proof, for it is a little deep) that cos 20°
satisfies no equation of degree 1 or degree 2 with integer coefficients.
Thus cos 20°, like �2, is an algebraic number of degree 3, and so by
the Theorem on Geometric Constructions, cos 20° is not constructible.
Thus the trisection of the 60° angle is impossible by straightedge and
compass methods.
Finally, consider the problem of squaring the circle. Given any circle,
we may consider its radius as the unit of length. With this unit, the area
of the circle is 11" square units. A square of equal size would have a side
length of vi;. So the problem of squaring the circle amounts to con­
structing a line of length vi; from a given unit length. Now it is well
78
N U M BE RS :
RATIONAL AN D
I RRATIONA L
known in geometric construction theory that from line segments of
lengths I and a, a line segment of length a 2 can be constructed. So if a
line segment of length v'; could be constructed, so also could a line
segment of length 11".
But we asserted in the preceding section that 11" is a transcendental
number, that is to say, 11" is not an algebraic number. Hence by the
Theorem on Geometric Constructions, it is not possible to construct
a line segment of length 11". Thus the "squaring the circle" construction
is impossible.
Problem Set 19
(Problems 2 and 3
structions. )
are for students with a knowledge of geometric con­
I . Prove that the first, second, and fourth numbers in the list ( 10) are roots
of equations ( 1 1), ( 1 2), and ( 1 4) respectively.
2. Prove that, given line segments of length 1 and length sin 20°, a line
segment of length cos 20° can be constructed by the straightedge and
compass methods.
3. Prove that, given line segments of length 1 and length tan 20°, a line seg­
ment of length cos 20° can be constructed by the straightedge and compass
methods.
5.6 Further Analysis of vi
I n the preceding section, we asserted that V"2 is an algebraic number
of degree 3. That is, V"2 is a root of the equation x 3
2 = 0, but it
is a root of no equation of degree I or degree 2 with integer coefficients.
We now prove this assertion.
To establish that V"2 is a root of no equation of degree I with integer
coefficients, we must prove that there are no integers a and b, with a
not equal to zero, so that
-
a V"2 + b
=
0.
If there were any such integers, then we would have V"2 =
bla, so
that V"2 would be a rational number. But we know that �2 is irrational,
by Corollary 2 of Section 4.3.
It is more difficult to prove that V"2 is not a root of any quadratic
equation with integer coefficients, such as
-
ax 2 + bx +
c =
0.
TR I G 0N O M ET R I C N U M BERS A N D LOG S
79
We presume that �2 is a root of such an equation, and then we deduce
a contradiction. Thus we presume that
a( �2)2 + b �2 + e
i.e., that
a �4 + b �2
=
=
0,
- c.
Squaring both sides and simplifying, we obtain
b2 �4 + 2a2 �2
=
e2 - 4ab.
The last two equations can be thought of as two simultaneous linear
equations in the quantities �4 and �i. Either they can be solved or
they cannot, depending on whether the pairs of coefficients, a,b and
b2,2a2, are not, or are, proportional.
If they can be solved, for example by eliminating �4, we get
v2
_3/-
=
4a2b - ae2 - b2e .
b 3 - 2a 3
Rut �2 is irrational, so we have a contradiction.
The other possibility is that the pairs of coefficients are proportional.
This means that
a
b
=
�2 =
b2
'
2a 2
b
a
_
.
Again we have a contradiction, and so we conclude that �2 is an
algebraic number of degree 3.
Problem Set 20
1.
2.
Prove that V2 is an algebraic number of degree 2.
Prove that {/i is an algebraic number of degree 3.
5.7 A Summary
In this chapter we have applied the methods developed earlier to
show that trigonometric functions and common logarithms have
irrational values in most of the cases listed in elementary tables. Then
we divided the real numbers into two new classes, the algebraic and the
80
N U M BE R S : RATION A L A N D I R RATION A L
transcendental numbers, and saw how these classes were related to the
earlier separation of real numbers into rational and irrational numbers.
Without proof, we learned that if a length can be constructed from a
given unit length by straightedge and compass, then that length is an
algebraic number ofdegree 2k, where k is some non-negative integer.
(A reader who is quite familiar with analytic geometry can persuade
himself somewhat of the truth of this theorem regarding geometric
constructions by analyzing the algebraic meaning of the steps which
can be performed with straightedge and compass. The three cases are
a straight line intersecting a straight line, a straight line intersecting a
circle, and a circle intersecting a circle.) Having thus eliminated the
possibility of constructing a straight line whose length is an algebraic
number of degree 3, we have seen that a cube cannot be doubled, nor a
general angle trisected, by the prescribed methods. Because of the
impossibility of constructing any straight line whose length is a transcen­
dental number, we have seen how the problem of squaring the circle
was settled : this construction too is not possible.
C H A P T E R S I X
The Approximation of Irrationals
by Rationals
This chapter is concerned with the closeness of approximation of an
irrational number by rational numbers. Now, as we shall see, we can
get rational numbers that are as close to v2, for example, as we
please. There are rational numbers alb which are within 10- 1 0 of v2,
or within 10-20 of v2, or within any other range of difference that we
want to specify. And this is true for any irrational number, not only
for v2.
But in order to find a rational number alb which differs from an
irrational number by less than 10-20, we must look for an alb with a
very large denominator b. If we allow b to be as large as 10 20, we can
find a fraction alb to meet the specifications. What happens if we
restrict b to be no larger than 1015 or 101 0 ? The problem becomes
deeper and more troublesome. In looking at questions of this kind, we
will be concerned with what can be said for every irrational number,
not j ust for some special cases like v2 and v3.
To talk about the approximation of one number by another, we
must have available the language and notation of inequalities. There­
fore, we shall begin with this topic.
6.1 Inequalitiest
u
We write u > v whenever u is greater than v, and this means that
v is positive. Of course, whenever u is greater than v, it follows
-
t For a thorough treatment of inequalities, see E. Beckenbach and R . Bellman,
An Introduction to Inequalities, in this series.
81
82
N U MBERS :
RATIONAL AN0
IRRATIONA L
that v is less than u, written as v < u in mathematical notation. Conse­
quently the four inequalities
u > v,
u - v > 0,
v < u,
v-u<O
are simply four ways of stating the same basic relationship between u
and v. Similarly, u � v means that u is greater than or equal to v, and
this amounts to saying that u - v is positive or zero, but not negative.
THEOREM 6. 1 .
(a) If u > v, and w is any number, then u + w > v + IV.
(b) If u > v, and w is any number, then u - w > v - w.
(c) If u > v, and w is any positive number, then uw > vw.
(d) If u > v, and w is any positive number, then u/w > v/w.
(e) If u > v, and if u and v are positive, then u 2 > v2 but l/u < l/v.
(f) If u > v, and v > w, then u > w.
(g) A ll these results hold if the inequality signs, > and < , are replaced
by � and � throughout.
PROOF : We shall take two principles for granted : that the sum and
the product of two positive n umbers are positive.
(a) We are given that u - v is positive, and we are to prove that
(u + w) - (v + IV) is positive. This is clear, because
u- v
=
(u + IV) - (v + IV).
(b) Again we are given that u - v is positive, and we are to prove that
(u - IV) - (v - w) is positive. As before, this follows because
u- v
=
(u - w) - (v - IV).
(c) We are given that u - v and w are positive, and we are to prove
that uw - vw is positive. This follows from the facts that uw - vw =
w(u - v), and that the product of two positive numbers is positive.
(d) This is really contained in part (c), because if IV is positive, so
is l/w. Consequently, l/w could be used as the multiplier in part (c) in
place of w, thus : if u > v, then u(l/w) > v(l/w) .
(e) Since u and v are positive, so is u + v. But u > v implies that
u - v is also positive, and hence that the product (u + v) (u - v) is
positive. Thus we have
(u + v) (u - v) > 0,
u2 - v2 > 0,
u2 > v2•
A PPROXI M A TI O N B Y R A T I O N A LS
83
On the other hand, if we use part (c) to j ustify the multiplication of
both sides of u > v by l /uv, we get
1
1
u·- > v·uv
uv
and hence
1
1
- >­
v
u
1
or
-
u
1
< -.
v
(f) We are given that u - v and v
w are positive, and we want to
prove that u - w is positive. However,
-
u- w
=
(u - v) + (v - w),
so once again we have merely to use the principle that the sum of two
positive n umbers is positive.
(g) One way to prove part (g) would be to run through parts (a),
(b), . . . , (f), giving a fresh analysis in each case. However, there is an
easier way. The proofs of parts (a) to (f) have been based on the princi­
ples that the sum and the product of two positive numbers are them­
selves positive. Now, whereas u > v means that u - v is positive,
u � v means that u - v is positive or zero, that is to say, u - v is non­
negative. Furthermore, the sum and the product of any two non-nega­
tive n umbers are themselves non-negative, and on this basis it must be
that all the proofs (a) to (f) will automatically extend from the > case
to the � case.
If the n umbers u and v are plotted on the real line, as explained in
Chapter 3, the inequality v < u has the significance that v is to the left
of u, or u is to the right of v (Fig. 1 8). The inequalities w < v < u are
v
-2
U
-1
I
o
1
2
Figure 1 8. Illustration of v < u
taken to mean that w < v and v < u, so that v is between
19). We must explain the use of the word "between."
w
I
and u (Fig.
u
I
v
I
Figure 19. 1lIustration of w <
IV
v
<
/I
84
N U M BE RS : RATIONA L A N D I RRAT I O N A L
If we write w < v < U, we mean v is "strictly between" Ii' and U and
may not coincide with either w or U; but if we say "between" or write
IV ;;;; v ;;;; u, we include the possibilities of v being equal to w or to u.
We may want to admit only one of these possibilities, for instance
w < v ;;;; U or IV ;;;; V < u. The symbols make the meaning perfectly clear.
Problem Set 2 1
I . Given that u 2 > v 2 and that u and v are positive, prove that II > v.
2. Given that r > s, prove that - r < -So
3 . Prove that terms in inequalities can be transposed from one side to
the other provided the signs are changed. Specifically, prove that if
a + b - c > d + e - /, then a - e + f > d - b + c.
4. G iven positive integers n and k such that n � k, prove that
I
I
;; � k
and
I
112
�
I
kn
'
5. Determine the truth or falsity of the following:
(a) If r > s, then r2 > s2.
(b) If r > s, and c is any number, then cr > cs.
(c) If - 1 /2 < � < 1 /2 holds, then - I < � < I holds. t
( d ) If - 1 /2 < � < 1 /2 holds, then - 3/2 < � < 3/2 holds.
(e) If 0 < � < 1 /2 holds, then - 1 /2 < � < 1 /2 holds.
(f) If - 1 /2 < � < 1 /2 holds, then - 1 /3 < � < 1 /3 holds.
(g) If - 1 /2n < � < 1 /2n holds, then - I /n < � < I /n holds.
6. A certain irrational number � lies strictly between - 10 and 1 0. Write this
in mathematical notation.
7. If w is negative and u > v, prove that uw < vw.
8. (a) Let u and v be any two different integers chosen from J , 2, 3, ' " , 1 0.
Prove that -9 � u - v � 9.
(b) If in (a) we did not prescribe that the integers u and v be different.
would it still be true that -9 � II - V � 9?
6.2 Approximation by Integers
If we round off any real number by replacing it with the closest
integer, the error committed will be at most 1/2. For example, if we
replace 6.3 by 6, or 9.7 by 10, or 7.5 by either 7 or 8, the error is no
more than 1 /2 in each case. If we replace an irrational number by the
nearest integer, the error is less than 1/2, and we begin the theory of
approximations with this simple case.
t The symbol >. is a Greek letter, read "lambda."
A PPROX I M ATION BY RATIONALS
85
THEOREM 6.2. Corresponding to any irrational number a there is a
unique integer m such that
1
1
- - < a - m < -.
2
2
PROOF. We choose m as the integer closest to a. For example, if
v) = 1 .73 · . " we choose m = 2, or if a = 2 v) = 3.46 · . " we
choose m = 3. Thus m may be the integer j ust larger than a, or j ust
smaller than a, whichever is closer. (It is clear that one of them is closer
to a than the other, otherwise a would be halfway between two con­
secutive integers, say n and n + 1 . But then a would equal n + 1 /2
which is a rational number, and this contradicts our assumption.) We
can state this in another way. Any line segment A B of unit length (i.e.,
one unit long) marked off on the real line, as in Fig. 20, will contain
a
=
-3
I
-2
I
-1
I
A
I
I
B
2
0
I
I
3
I
Figure 20
exactly one integer unless A and B happen to be integer points. Now
take A to be the point corresponding to the number a - 1 /2 and B
corresponding to a + 1 /2. Since a - 1 /2 and a + 1/2 are not integers
(they are not even rational-see Theorem 4 . 1 , Chapter 4), we know
that A and B cannot be integer points. Denoting the unique integer in
the segment A B by m, we see that m lies strictly between a - 1 /2 and
a + 1/2. Thus
a
-
1
1
- < m < a + -·
2
2
Subtracting a, we get
1
1
- - < m - a < -.
2
2
Now if a number m - a lies between - 1/2 and 1 /2, so will the number
obtained by changing the sign, and hence a - m also lies between
- 1/2 and 1/2. Thus we get the inequalities of Theorem 6.2.
The integer m is unique, because if there were another integer n
satisfying
1
-- < a
2
-
1
n < -,
2
N U M BERS : RATIONA L AN 0 I RRATIONA L
86
then n would also satisfy
1
-- < n
2
Adding
a
-
a
1
< -.
2
to these inequalities, we see that n would satisfy
a
-
1
- <n <
2
1
2
+ -.
a
But the segment A B contained only one integer, so n is the same as m.
Problem Set 22
may wish to use
3 . 1 4 1 59 · . . . )
(In this and subsequent problem sets, the reader
vi = 1 .4142 1 . . . , v3 = 1 .73205 · · . , 7r
1 . Find the integers closest to
=
(a) v2,
(b) 2vi,
(c) 3 vi,
(d) 4v2,
([) 4v3,
(g) 7r,
(h) 107r,
(i)
2. Given any irrational number
that 0 < a
q < 1.
a,
-
v3,
(e) 3V3
(j)
-
77r
prove that there is a unique integer q such
-
6.3 Approximation by Rationals
One way to approximate an irrational number, such as v2, is to use
the decimal form
v2
=
1 .4142 1 . . . .
The numbers 1 , 1 .4, 1 .4 1 , 1 .414, 1 .4142, 1 .4142 1 , . . . form a series of
closer and closer approximations to v2. These approximations are all
rational numbers, and so we have an infinite sequence of rational
approximations to v2:
(1)
14
l' 10
'
141
'
100
1414
14142
1000' 10000
'
14142 1
100000'
These numbers are getting closer and closer to v2 as we go further
along in the series. Furthermore, we can write the inequalities
A P P R OX I M AT I O N
BY RATION A L S
87
! < v2 < �,
1
1
14
15
v-2
< 1 0'
10 <
141
1 42
v 2< 100 '
100 <
1414
1415
v2
< 1 000'
1000 <
14142
14143
v2
< 1 0000'
10000 <
141421
141422
v2
< 100000' etc.
100000 <
These inequalities show that infinitely many of the terms of ( 1 ) are as
close to v2 as we want to specify. Suppose, for example, we want to
know that there are infinitely many rational numbers within 0.000 1
of v2. These we can get by taking all but the first four terms of the
sequence ( 1).
However, the rational numbers (1) have the special feature that their
denominators are powers of 10. There might be better approximations
to v2 among rational numbers in general, without any restriction on
their denominators.
Let us turn to the irrational number 7r for a well-known instance
that illustrates our discussion. Since 7r has the value 3 . 1 41 59 · " the
sequence for 7r analogous to ( 1 ) is
.
(2)
3
31
l' 10
'
314
'
100
3 141
'
1000
3 14 1 5
'
10000
3 14 1 59
'
100000
However, we know that 22/7 is a closer approximation to 7r than 3 1/ 1 0.
In fact 22/7 is closer than 3 1 4/ 100, although not closer than subsequent
terms of the sequence (2).
To get away from our dependence on the denominators 10, 102,
103, etc., we first show that every irrational number can be approximated
by a rational number having any given denominator.
THEOREM 6.3. Let � be any irrational number and n be any positive
integer. Then there is a rational number with denominator n, say min,
such that
m
1
1
�
--.
2n < - n < 2n
N U M BERS : RAT ION A L A N D I RRAT I0NA L
88
We motivate the proof of this theorem with an example. Suppose
that � is v2 and n is 23. Consider the irrational number 23 v2, which,
by use of the decimal expansion 1 .4142 1 . . . for v2, has the approxi­
mate value
23 v 2
=
32.52 . . .
Thus the closest integer to 23 v2 is 33, and this is the "m" of
Theorem 6.2 which, for ex = 23 v2, states that
1
1
- - < 23 v2 - 33 < - .
2
2
But 33 is also the "m" of Theorem 6.3 : for, according to Theorem 6. 1 ,
we can divide these inequalities by 23 to obtain
1
1
33
- - < v" - - < -.
23
46
46
PROOF. In general, beginning with any irrational � and any positive
integer n, we note that by Theorem 4. 1 of Chapter 4, n"X is irrational.
Then we define m as the nearest integer to n�, and so by Theorem 6.2,
1
1
- - < n� - m < -.
2
2
By Theorem 6. 1 , these inequalities can be divided by the positive integer
n to give
m
1
1
- - < � - - < -.
n
2n
2n
Hence Theorem 6.3 is proved.
Find rational numbers mIn as in Theorem 6.3 for the case where
= 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10.
SOLUTION. By simple calculation the integers nearest to
V2, for n
EXAMPLE.
�
=
� 2� 3� 4� 5� 6� 7� 8� 9� 1�
are 1, 3, 4, 6, 7, 8, 10, 1 1 , 1 3, 14. Hence the rational numbers to be determined
are
3 4 6 7 8 10 1 1 13 14
I' 2' )' 4' :5' 6' 7' 8' 9' 1 0'
APPROXIM ATION
BY
89
RATIONALS
and the error in each of these approximations is less than 1 /2n, where
integer in the denominator.
n is the
This example shows that the rational fractions min of Theorem 6.3
not necessarily in lowest terms.
a re
Problem Set 23
1. Find rational numbers mIn as in Theorem 6.3 for the case where �
and n = 1 , 2, 3, 4, 5, 6, 7, 8, 9, 1 0.
2. Find rational numbers mIn as in Theorem 6.3 for the
� = 7r = 3 . 1 4 1 59 · . . , for n = 1 , 2, 3, 4, 5, 6, 7, 8, 9, 1 0.
=
v3
case where
3. G iven any irrational number � and any positive integer n, prove that there
is an integer m such that
_l
n
• 4. For a fixed irrational number
< �
_�
n
<
l.
n
� and a fixed positive integer n, prove that
there is only one integer m that satisfies the inequalities of Theorem 6.3.
• 5. Prove that Theorem 6.3 would be false if the words "in lowest terms" were
inserted with reference to mIn, thus : " . . . Then there is a rational number
in lowest terms with denominator n, say mIn, . . . "
6.4 Better Approximations
Theorem 6.3 states that any irrational n umber A can be approximated
by a rational n umber min "to within 1/2n," i.e., with an error less than
1/211 . Can this be done to within 1/3n, or 1/4n, or perhaps closer? The
answer is yes. I n the next theorem we show that � c�n be approximated
by min to within l/kn for a ny k we wish to specify : k = 3, k = 4,
k = 1000, etc. But, whereas in Theorem 6.3 the approximation to with­
in 1/2n could be achieved for every positive integer n, the approximation
to within Ijkn with prescribed k in Theorem 6.4 cannot be obtained
for all n.
Can we approximate any irrational n umber � hy min to within l/n2,
or l/n3, or even closer? To within 1/n2, yes ; to within l/n 3 , no. But
these are the topics of later sections. So let us begin with approximations
of � by min to within l/kn.
THEOREM 6.4. Given any irrational number � and any positive integer
k, there is a rational number min whose denominator n does not exceed
k, such that
1
-- <
nk
� -
m
1
< -.
nk
n
N U M BERS : RATI O N A L A N D I R RATIONA L
90
Before presenting a proof of Theorem 6.4 that is valid for any � and
k, we shall prove the theorem for a particular instance, namely with
� = v3 and k = 8. First we enumerate the multiples of � from 1 . �
to k . �. We list the multiples of v3, writing each multiple as a sum of
two positive numbers, an integer and a number less than one :
v3 = 1 + 0.732 . . . ,
v3 - 1 = 0.732 · . . ,
2 v3 = 3 + 0.464 · · · ,
2 v3 - 3 = 0.464 · · · ,
3 v3 = 5 + 0. 1 96 · · . ,
3 v3 - 5 = 0. 1 96 · . . ,
4 v3 = 6 + 0. 928 . . ,
4
5 v) = 8 + 0.660 · · · ,
5 v) - 8 = 0.660 · · · ,
6 v) = 10 + 0.392 . . . ,
6 v) - 10 = 0.392 .
7 v) = 12 + 0. 124 · . . ,
7 v)
8 v) = 1 3 + 0.856 · . . ,
8 v) - 13 = 0.856 · . . .
.
v) - 6 = 0.928 · · · ,
-
.
12 = 0. 124 · .
.
,
"
The expressions in the right-hand column were obtained from those on
the left by subtracting the integer part.
Next, we separate the unit interval into eight parts, I" h, . . . , Is, as
11
I
.1
8
0
12
I
2
"8
13
I
:3
8
14
I
4
8"
Is
Figure 2 1
I
i
8
16
I
.§.
8
17
I
.2
8
18
I
� =1
shown in Fig. 21. Thus II consists of the numbers between 0 and 1/8,
h the numbers between 1/8 and 2/8, h the numbers between 2/8 and
3/8, and so on. t Then we classify the eight decimal parts of the mUltiples
of v3 into the categories It. Iz, . . . , Is as follows :
0.732 ·
0.464 .
0. 1 9 6 .
0. 928 .
0. 660 ·
0.392 .
0. 124 ·
0.856 .
. is in 1 6 (because 0. 732 · . . is between 5/8 and 6/8),
. is in 14,
. is in 1 ,
2
is in I s,
. is in 1 6,
. . is in 14,
. . is in I "
. . is i n 1 7 ,
.
.
.
.
.
.
t Since we wish to derive strict inequalities, it is convenient to interpret "between"
as "strictly between" ; thus the i ntervals Ij include all points /I such that
(j - 1 )/8 < /I < j/8 .
APPROXIMATION BY RATIONA LS
91
We use the number in the above list which is in II :
0. 124 ·
.
is in II ; that is, 7 v) - 12 is in II '
.
But the numbers in II lie between 0 and 1/8, so
I
o < 7 v3 - 1 2 < -
8
.
Now, since the number 7 v ) - 1 2 lies between 0 and 1 /8, i t certainly
lies between - 1 /8 and 1 /8, thus
1
1
- - < 7 v 3 - 12 < - .
8
8
Dividing this inequality by 7, we get
-
I
7·8
-
I .
12
< v3 - - <
7
7·8
-
This is a result of the form stated in Theorem 6.4, with k = 8, n = 7,
and m = 12.
Our argument was based on the fact that 7 v) - 12 was in II ' What
would we have done if there had been no number in the interval II ?
The answer is that if there had been no number in 11 0 then one of the
intervals h, h,
, 17 would contain two or more numbers. In the
present example, not only is there a number in 11 0 but also there are
two in 14 and two in 16, Consider the pair in 16 :
.
.
.
0.732 · . is in
16 ;
0.660 · . is in
16 ;
.
that is, v) - 1 is in h ;
and
.
that is, 5 v) - 8 is in
16,
Now, whenever two numbers are in 16 (or in any one of these intervals),
they are within 1/8 of each other, so their difference lies between
- 1/8 and + 1 /8. I n particular, for the two numbers in 16, we have
1
I
- - < (5 v 3 - 8) - ( V 3 - 1) < - ,
8
8
-
-
I
I
- - < 4 v3 - 7 < - .
8
8
92
N U M BERS : RATIONAL AND I RRATIONAL
Dividing by 4 we get
7
1
1 ,
- - < v3 - - < 4·8
4 4·8
and this is another result of the form stated in Theorem 6.4 for X = v3
and k = 8, this time with n = 4 and m = 7.
PROOF OF THEOREM 6.4. The special instances we have j ust discussed
can serve as models for the proof of Theorem 6.4. Given an irrational
number X and a positive integer k, we take the k numbers X, 2X, 3X,
4X, . . . , kX and write each of these numbers as an integer plus a frac­
tional or decimal part :
X = al +
2X = a2 +
3X = a3 +
4X = a4 +
/3 10
/32,
/33 ,
/34'
X
2X
3X
4X
-
-
-
-
a l = /3 10
a2 = /32,
a3 = /33,
a4 = /34,
The symbols ah a2, . . . , ak stand for integers, whereas the symbols
/3 10 /32, . . . , /3k stand for numbers between 0 and 1 . Next we divide the
unit interval into k parts, II > h . . . , h, each of length 11k (Fig. 22).
11
0
I
.!
k
12
1
I
�
k
3
I
2
k
14
I
�
k
I
k-l
k
. ..
Figure 22
1k
I
1.
k
: 1
Thus the interval II consists of the numbers between 0 and 11k, h the
numbers between 11k and 21k, h the numbers between 21k and 3/k,
etc. The word "between" is here used in the strict sense, so that, for
example, the numbers 21k and 31k are not themselves members of the
interval h Note that, by Theorem 4. 1 of Chapter 4, each of the
numbers /3 1 0 /32, . . . , /3k is irrational. Consequently no {:3 can equal any
of the rational numbers
0, Te' Te' Te'
1
2
3
... ,
k- 1
-- '
k
k
Te'
Thus each /3 lies in exactly one of the intervals 110 h 13, . . . , 'k·
A P P RO X I M A T I O N B Y R A TI O N A L S
93
There are two possibilities regarding II : either II contains one or
more of the /3's, or II contains none of the /3·s. We shall treat these two
possibilities separately.
Case 1. The interval 1 1 contains one or more /3·s. Thus there is one /3,
say /3n, in the interval II' The symbol n stands for some integer among
1. 2. 3,
k. The number /3n is the same as nX - an, so we know that
.
.
.
•
1
0 < nX - an < l(
because II is the interval from 0 to 11k. It follows that
- l( < nX - a" <
1
k'
1
and if we divide through by n we get
1
a"
1
- - < X - - < -.
kn
n
kn
Thus Theorem 6.4 is proved in this case, because we can define m to
be the integer an.
Case 2. The interval II contains none of the /3·s. In this case, the k
numbers lie in the k - 1 intervals
h. h,
. . . . h - I'
At this point we apply the Dirichlet pigeon-hole principle, which says
that if there are k pigeons in k - 1 holes, there must be at least one
pigeon-hole with two or more pigeons in it. Thus there must be at
least one interval containing two or more /3·s. Let us say that /3r and
/3j are in the same interval, where r and j are two different numbers
among 1, 2, 3, . . . , k. Presuming that j is larger than r, we then know
that j - r is a positive integer less than k.
Since /3r and /3j lie inside the same i nterval of length 11k, their
difference lies between - 11k and 11k. Thus
But /3j
=
j - aj and /3r
=
r - ar, so that
i < (jX - aj) - (r - ar < i
-i < (j r)X - (aj - ar) < i'
-
or
)
-
94
N U M B E RS : R AT I O N A L A N D I R RA T I ON A L
Define n as .i -
r,
and m as aj - ar, and then we have
1
1
- - < nX - m < - .
k
k
We see that n is a positive integer by definition, so by Theorem 6. 1(d)
we may divide by n and obtain
1
1
m
- - < X - - < -.
kn
kn
n
Futhermore, we know that since n is equal to j - r, it is less than k,
and so the proof of Theorem 6.4 is complete.
Note that the number min is not necessarily in lowest terms. If i r
and aj - ar have no common factor, min is in lowest terms ; otherwise
it is not.
-
Problem Set 24
I . Following the statement of Theorem 6.4, there is an example given of the
instance X = v3" and k = 8. What values of m and n would have resulted
if we had singled out the two numbers 0.464 . . . and 0.392 . . . in 14,
instead of the two numbers in h?
2. Apply the method given in the proof of Theorem 6.4 to each of the following
cases, and so get values of m and n satisfying the inequalities of Theorem
6.4 :
(a) X =
(b) X
=
(c) X
=
(d) X
=
(e) X =
(0 X
=
(g) X
=
v3",
v'3,
v'3,
v3",
VZ,
VZ,
vz,
k
=
2;
(h) X
=
k
=
4;
(i) X
=
k = 6;
(j) X
=
k
=
10;
(k) X =
k
=
2;
(I) X
k
=
4;
(m) X
k = 6;
(n) X
=
VZ,
vz,
vi,
k
=
8;
k = 10;
k
=
14;
11",
k
=
2;
=
11",
k
=
4;
=
11",
k
=
6;
11",
k
=
8.
6.S Approximations to Within l /n2
At the beginning of Section 6.4 we indicated the direction of our
study, namely to try for better approximations of any irrational number
X. From the approximation of X by min to within 1/2n for any n in
A PP R O X I M ATION B Y R A TI O N A LS
9S
Theorem 6.3, we moved to approximation to within llkn for some
n � k in Theorem 6.4. Now we obtain approximations to within Iln2•
THEOREM 6.5. Given any irrational number X, there are infinitely many
rational numbers min in lowest terms such that
1
-- < X
n2
-
m
1
- < _.
n2
n
PROOF. First we observe that any rational number min satisfying the
inequality of Theorem 6.4 automatically satisfies that of Theorem 6 . 5 .
The reason for this is that since n does not exceed k, from k � n we
may deduce, using Theorem 6. 1 , parts (d), (e), and (g), that
1
1.
< kn = n2
and
-
Hence any number which lies between - llkn and llkn must certainly
lie in the range between - lln2 and Iln 2 •
Next, we show that if any rational number min, not in lowest terms,
satisfies the inequalities of the theorem, then the same rational number,
in lowest terms, must also satisfy the appropriate inequalities. Let us
write MIN as the form of min in lowest terms. We may presume that
both n and N are positive, any negative sign being absorbed into the
numerator. Hence we have
0 < N < n,
because the reduction to lowest terms does not alter the value of the
fraction but does reduce the size of the denominator. It follows from
Theorem 6. 1 that
1
-1 < ­
n
N
and
1
1
<
n2
N2'
and so if X satisfies
1
-- < X
n2
-
1
m
- < -,
n2
n
96
N U M BERS : RATIONAL A N D I RRATIONAL
then it automatically satisfies
1
-- < X
N2
-
M
1
- < --N2
N
In order to complete Theorem 6.5, all we must prove is that there
are infinitely many rational numbers min, in lowest terms, that satisfy
the inequalities. Suppose, on the contrary, that there were only a finite
number of these fractions, say
Then consider the i numbers
...,
These are all irrational by Theorem 4.1 of Chapter 4, and so none
of them is zero. Some may be positive and others negative, and we
choose an integer k so large that 11k lies between 0 and all the positive
numbers, and also so that - I lk lies between 0 and all the negative
n umbers. We can do this because the larger we choose k, the closer
11k and - 11k get to O. Thus we have chosen k so large that all the
following inequalities are false :
1
- Ie
(3)
ml
< X
1
-- < X
k
-
-
�
<
1
r
m2
1
- < -,
n2
k
1
1
mj
- - < X - - < -.
k
nj
k
With this value of k, we apply Theorem 6.4 and obtain a rational
n umber min such that
m
1
1
- - < X - - < -.
kn
n
kn
97
APPROX I M ATION BY RATIONALS
Now, this says that X
min lies between - l/kn and I/kn, and so
X min must lie between - Ilk and 11k ; in symbols,
-
-
I
-- < X
k
-
m
1
- < -.
k
n
But since all the inequalities (3) are false, we conclude that min is
different from each of the i numbers m t !n l o m21n2 ' . . . , m;fni. Therefore
we have obtained one more rational fraction to satisfy the inequalities
of Theorem 6.5.
EXAMPLE. Find four rational approximations (in lowest terms) to the
irrational number 11", close enough to satisfy the inequalities of Theorem 6.5.
SOLUTION.
First we observe that since 11" =
I
3
I
11"
-12
1 2 < - -I < -
3. 1 4 1 59 · . .
I
,
4
I
-( 2 < 11" - -I < p.
and
To find two others, we can use the method of Theorem
rational numbers with denominators 2, 3, and so on :
6.3 to get the closest
2' ]' 4' S' 6' 7' . . . .
6
9
13
16
19
22
We reject 6/2 and 9/3 because they are not in lowest terms, and we test the
others in the inequalities of Theorem 6.5 ; for example,
-
I
36 <
11"
-
19
"6 < 36·
I
(True !)
Thus we are led to reject 1 3/4 and 1 6/5, and accept 1 9/6 and 22/7. So one
set of answers to the question would be 3/1 , 4/1 , 1 9/6, and 22/7.
The rational number 22/7 is a very good approximation of 11". There is no
rational number with denominator between 1 and 56 that is closer to 11".
The rational number 1 79/57 is slightly closer to 11" than 22/7, but it does not
satisfy the inequalities of Theorem 6.5. The rational nU!Jlber 355/ 1 1 3 satisfies
the inequalities of Theorem 6.5 and is markedly closer to 11" than is 22/7.
In fact it is accurate to six decimal places.
It is possible to prove the following stronger version of Theorem 6.5 :
Given any irrational number X, there are infinitely many rational numbers
min in lowest terms such that
1
< X
n(n + 1)
-
m
n
<
n(n + 1)
98
N U M BERS : RATI ON A L AN D I RRAT I0N A L
With the help of this theorem, the number 4/1 (which is a relatively
poor approximation to 11'") can be eliminated in the above example.
In order to prove the stronger version of Theorem 6.5, we need a
stronger version of Theorem 6.4. We shall sketch only the main steps
and leave the details to the reader.
In the proof of Theorem 6.4, Dirichlet's pigeon-hole principle Was
used to argue that, given k numbers distributed over k intervals, either
there is a number in the first interval or there is an interval containing
at least two of these numbers. To get the stronger version of Theorem
6.4, we divide our unit interval into k + 1 subintervals and argue :
Given k numbers distributed over k + 1 intervals, either there is a
number in the first interval, or there is a number in the last interval,
or there exists an interval containing at least two numbers. This use
of the pigeon-hole principle enables us to substitute the stronger
inequality,
1
X
n(k + 1) <
m
-
n < n(k + 1)'
1
for that now appearing in Theorem 6.4, without otherwise changing
the statement. The proof of the stronger version of Theorem 6.5 is
now immediate.
Problem Set 25
I . For a given irrational X, prove that two of the "infinitely many rational
numbers min" of Theorem 6.5 have n = I , i.e., are integers.
2.
Let X be a given irrational number. Prove that, apart from one exception,
any rational number that satisfies the inequalities of Theorem 6.5 auto­
matically satisfies the inequalities of Theorem 6.3.
3. Find two rational numbers, which are not integers, that satisfy the inequal­
ities of Theorem 6.5 for
(a)
X
=
v'2;
(b)
X
=
vi);
(c)
X
=
vis.
4. (a) Of the first five numbers in the sequence ( I ), which ones satisfy the
inequalities of Theorem 6.3, with X
viZ?
(b) Which ones satisfy the inequalities of Theorem 6.5 ?
=
5. (a) Of the first five numbers in the sequence (2), which ones satisfy the
i nequalities of Theorem 6.3, with X = vi)?
(b) Which ones satisfy the inequalities of Theorem 6.5?
• 6. Prove that the statement of Theorem 6.5 is false in case X
=
3/ 5 .
• 7. (a) Let alb and min be rational numbers, in lowest terms, with positive
A PPROXI M ATION B Y RATION A LS
99
denominators. Prove that they are unequal if n > b. Hence prove
that for n > b the inequalities
a
m
i
1
-bn < 'b - -; < bn
are false.
(b) Prove that the statement of Theorem 6.5 is false if X is any fixed rational
number, say X = a/b .
• 8. Complete the proof of the stronger version of Theorem 6.5 (following the
sketch given j ust prior to this problem set) ; show that 71' - 4/ I and
1 9/6 do not satisfy the stronger inequality, but that 71' - 22/7 does.
71'
-
6.6 Limitations On Approximations
We proved in Theorem 6.3 that, corresponding to any irrational
number X, there are infinitely many rational numbers min such that
m
1
1
-- < X - - < - .
2n
n
2n
Then, in Theorem 6.S we established that there are infinitely many min
such that
1
1
m
-- < X - - < - .
n2
n
n2
Is it possible to prove that there are infinitely many min such that
-
1
1
m
< X--<
?
2n2
n
2n 2 '
-
-
The answer is yes, although we shall not prove it here. In fact, there is
a famous theorem which states that there are infinitely many min
corresponding to any irrational number X such that
1
m
1
- --- < X - - < -v'S n2
n
-
v'S n2
,
and furthermore that v'S is the constant which yields the best possible
approximation of this kind. This means that if v'S is replaced by any
larger constant, the statement becomes false.
To give some idea as to how it is possible to prove that there is a
limit on the size of the constant, we establish the following result :
1 00
N U M BERS : RATION A L A N D I RRATION A L
There are not infinitely many rational numbers min such that
1
V2
<
5n2
(4)
_ _
_
�
1
< _.
5n2
n
In fact we prove that (4) is impossible for any integer n greater than 10.
The proof is indirect. We assume that (4) holds for some integers m
and n, with n > 10. The inequality
1
V2
<
5n2
_ _
_
�
n
implies, for n > 10, that
(5)
11 < v 2 +
m
. r-.
1
1
< v2 +
< 2.
5n2
500
On the other hand, the inequality
r-.
m
1
v2 - - < n
5n2
implies, for n > 10, that
(6)
m
,r;;;
- > v2
n
.
-
1
- > v2
5n2
-
1
- > 1.
500
Now, if we add min to the members of the inequalities (4), we get
(7)
1
1
,r;;; m
m
- - - < v2 < - + - ·
5n2
5n2
n
n
According to Theorem 6.1(e), each of these three parts can be squared
and the inequalities retained, provided we prove that each part is
positive. By (6) we see that
m
i
l
1
---> 1 --> 1 - -> 0
5n2
5n2
500
'
n
Hence all parts of (7) are positive, and so we square throughout to
get
A PPROX I M ATl0N BY RAT l0N A LS
(m- - -) 2 < 2 < (m- + ) 2
n
5n2
n
5n2 '
1
101
1
-
m2
2m
1
m2
2m
1
- - - + - < 2 < - + - + -.
n2
5n3
25n4
n2
5n 3
25n4
M Ultiplying by n2, we get
2m
1
2m
1
m2 - - + - < 2n2 < m 2 + - + -- .
5n
25n2
5n
25n2
(8)
Now, by (5), we see that
m2 +
(9)
()
2
2 m
1
1
< m 2 + (2) +
+
"5 n
"5
25n2
25n2
1
< m2 + 4 +
< m2 + 1 .
"5
2500
On the other hand, by (5), we can write
( 10)
m2 -
Km-) +
5 n
1
25n2
-
()
4
2 m
> m2 - - > m 2 - 1 .
5
5 n
> m2 - - -
Applying (9) and ( 1 0) to (8), we obtain
()
2 m
1
2m
1
m2 - 1 < m2 - - - + -- < 2n2 < m2 + - + -- < m2 + 1
'
5 n
25n2
5n
25n2
m 2 - 1 < 2n2 < m2 + 1 .
But 2n2 i s a n integer, so if it lies between the integers m 2 - 1 and
m2 + 1, it must equal m2• Hence we conclude that
2n2 = m2,
m2
2 = -,
n2
v/2
=
'!!.,
n
and this is a contradiction, since v/2 is irrational, while m and n were
assumed to be integers.
\ 02
N U M BERS : RATIO N A L A N D IRRATIO N A L
Problem Set 26
I . (a) Prove that there are no rational numbers min, with n
_ _1_
<
vi
<
v'2
_
�
<
�
<
> 1 0,
such that
_1_ .
5n2
5n2
n
(b) Find all rational numbers min satisfying these inequalities.
2. (a) Prove that there are no rational numbers mjn, with n > 1 0, such that
�
-
L
n
n3
n3
(b) Find all rational numbers mjn satisfying these inequalities.
3. (a) Prove that there are no rational numbers mjn, with n > 10, such that
m
1
1
_ r
-- < v 3
- < -.
n
n3
n3
(b) Find all rational numbers satisfying these inequalities.
-
-
6.7 Summary
We have established several results about how closely any irrational
number X can be approximated by infinitely many rational numbers
min. The strongest theorem asserted that X can be approximated to
within l/n 2 • Then in Section 6. 6 we established a negative conclusion,
namely that there do not exist infinitely many rationals min within
1/{5n2) of V2 . A similar negative conclusion applies to any algebraic
number. It is true, but not proved here, that for any algebraic number
X, there do not exist infinitely many rational numbers min within l/n 3
of X. This cannot be said of transcendental numbers in general ; it is
true of some, but not all, transcendental numbers. In the next chapter,
we shall exhibit a number which can be approximated by infinitely
many min not only to within l/n 3 but to within l/n4, l/n 1 00, and,
indeed, to within l /nj for any j the reader cares to name, however large.
It will be proved that the n umber so exhibited is not algebraic, and thus
we will have shown that there are such things as transcendental
n umbers. Up to now we have spoken of them without knowing that
they even exist !
C H A PT E R S E V E N
The Existence of Transcendental
Numbers
How do we know that transcendental numbers exist? We shall
answer this question in this final chapter. It is easy enough to exhibit
a transcendental number ; to prove that it is transcendental is quite
another matter. The specific number whose transcendence we shall
establish has the important feature that its decimal expansion consists
mostly of zeros. We shall denote it by Of, and its value is
a =
0. 1 1 000 1 0000 . . . ,
where the ones occur in the decimal places numbered
1,
2,
6,
24,
120, 720,
5040,
... ,
that is to say, the decimal places numbered
l !, 2 ! ,
3 !,
4!,
5 !,
6 !,
7!,
The symbol k !, where k i s a natural number, i s read k factorial and
denotes the product of all the natural n umbers from 1 to k ; thus
k!
=
1 · 2 · 3 · . . . . (k - 2) . (k - 1 ) . k .
All the digits in the decimal expansion of a are zero except those
described above in the factorial n umber positions. Consequently a can
be written as a sum of negative powers of 10; that is,
(I)
=
1O- 1 ! + 1O- 2 ! + 1O-3! + 1O- 4 ! + 10-5 ! + . . .
10-1 + 10-2 + 10- 6 + 10-24 + 10- 1 20 + . . .
=
0. 1 + 0.01 + 0.00000 1 + . . . .
a =
103
1 04
N U M BERS : RATIONA L A N D I RRATIONAL
This number a is called a Liouville number, after the French mathe­
matician who first demonstrated that transcendental numbers exist.
What concrete property of the transcendental number a can we use
in order to prove that a is not algebraic ? The answer is that a can be
approximated by infinitely many rational numbers min, not only to
within 1/n 2 (this could be done for any irrational number, see Chapter 6)
but to within 1/n 3 , 1/n4, and, in fact, to within l/nr, where r is any
positive number whatever. No algebraic n umber has this property. If
>. is any irrational number, it can be approximated to within 1/n2 by
infinitely many rational min, as we saw in Theorem 6.5. But if >. is
algebraic, it cannot be approximated by infinitely many min more
closely, not to within 1/n 3 , or even to within 1/n2 . 1 ; within 1/n2 is the
best possible among all l/nr. To find this kind of a result about alge­
braic numbers was for many years an outstanding unsolved problem.
It was settled in 1955 by the British mathematician K. F. Roth who,
for this ingenious work, was awarded a Fields medal in 1958 at the
International Congress of Mathematicians in Edinburgh, Scotland.
The result is known as the Thue-Siegel-Roth theorem because A. Thue
and C. L. Siegel proved certain underlying results upon which Roth's
work was based.
As we said, proving the transcendence of a is a more difficult matter
than merely writing the decimal expansion of a. We shall use the ideas
on inequalities from Sect. 6. 1 . We also need the concept of absolute
value. Perhaps the reader is familar with this concept, but in case he is
not, we give a brief introduction to this topic and also to the factor
theorem.
7.1 Some Algebraic Preliminaries
Any real number a is either positive, negative, or zero. For every
s uch number a, we shall define the "absolute value of a," denoted by
the symbol lal. t If a is positive or zero, we define the absolute value of
a by the equation lal = a. If a is negative, the definition is lal = - a.
For example,
1 0 1 = 0,
1 3 1 = 3,
171
=
7,
1 -4 1 = 4,
I - 5 1 = 5,
1 - 61
=
6,
1 - 1000 1 = 1000 .
I nstead of separating the definition into cases in which a is positive,
t For a detailed treatment of the concept of absolute value, see Chapter I I I of
the monograph by Edwin Beckenbach and Richard Bellman, in this series.
EX I STENCE OF TRANSCENDENTAL N U M B E R S
105
zero, or negative, we could define the absolute value of a by the single
equation
la l = VOi,
(2)
,
because of the convention that VOi never designates a negative value.
One basic result is that if two numbers are equal, so are their absolute
values. In symbols, if a = b, then l al = I b l . Another simple consequence
of our definition (2) is that a and -a have the same absolute value,
whatever may be the value of a . In symbols, l a l = I -al.
Another important result is that labl = l a l . I bl . We can prove this
easily by using (2), as follows :
l a l = VOi,
Ib l = Ybi,
labl
=
v'a2b2
= VOi . Ybi,
so that
labl
=
lal . Ib l ·
Next, how is l a + b l related to the sum l a l + Ib l ? We shall show
that l a + bl � lal + Ibl. To prove this result, known as the triangle
inequality in the wider setting of complex n umbers, we separate the
problem into cases. If a and b are both positive, then
lal
=
a,
I b l = b,
so that
l a + b l = l al + I b l ·
If a and b are both negative, then
la + b l
=
-a -b,
la l
=
- a,
I b l = - b,
so that, again,
l a + b l = la l + I b l ·
If a and b are of mixed signs, one positive and the other negative, then
a + b involves, in effect, some cancellation, and we can say that
la + bl is less than the larger of l a l and Ibl.
It follows that la + b l < l a l + I b l .
106
N U MBERS : RATIO N A L A N D IRRATIONA L
If one of the numbers is zero, for example if b = 0, then
l a + b l = l a + 01 = l a l ,
I h l = 1 0 1 = 0,
so
la + b l = l a l + I b l ·
In summary, we see that in all cases, we have either
l a + b l = la l + I b l
or
la + b l < lal + Ib l ·
All these results on absolute values are collected in the following
theorem for convenience.
THEOREM 7. 1 . For all real numbers a and b, we have:
(1)
(2)
(3)
(4)
If a = b, then l a l = I b l ;
l a l = I - al ;
l ab l = l a l . I b l ;
la + bl � lal + I bl .
Next, we prove the factor theorem from algebra. Actually we prove
it in a special setting for our own later purposes.
THEOREM 7.2. Let f(x) he a polynomial with integer coefficients, and
let the rational number (3 be a root off(x) = O. Then x
(3 is afactor of
f(x); that is, there is a polynomial q(x) such that Jtx) = (x
(3) q(x).
Furthermore, q(x) has rational coefficients, and has degree one less than
the degree off(x).
-
-
PROOF. If we divide x - (3 into f(x), there results a quotient q(x) and
a remainder, say r. Since the degree of the remainder is always less than
that of the divisor (which, in our case, is the first degree polynomial
x
(3), we see that r is a constant independent of x. It follows that
-
f(x) = (x - (3) q(x) + r,
and since the steps in the division process are so-called rational opera­
tions, we see that q(x) must have rational coefficients. The equation
above is an identity in x, so we can now replace x by (3 to getf({3) = r.
However, f({3) = 0 because (3 is a root off(x) = O. Hence r = O. Thus
the divisionJtx) by x - (3 gives a zero remainder, and sof(x) = (x - (3) q(x).
Finally, whatever the degree off(x), we see that q(x) has degree one less.
EX 1STE N eE 0F TR A N seEN D E NT A L N U M BER S
107
Problem Set 27
I . Write the values of 121, 1 - 21, 1 -81, and 1 10-1 1 .
2 . I n the text it was established i f a = b , then lal = I b l . I s the converse true?
3. Prove that la + b + cl � la l + Ibl + lei .
4. (a) Prove that I x + 7 I = x + 7 if x � - 7, but I x + 71 = - x - 7 if
x � - 7.
(b) Give a similar analysis of I x - 71 = x - 7.
5. For what values of x, if any, do the following equations hold?
(c) I x + 7 1 + I x - 7 1 = I x l + 7 ;
(a) I x + 71 = 5 + Ix l ;
(b) I x l = I x - 4 1 ;
(d) 12xl = 2 Ix I 6. Prove that the inequalities of Theorem 6.5 of Chapter 6,
n2 <
-�
can be written in the form
>.
-
n
�
< n2
�,
I>. - �I < L
n
n2
7. Prove that 8 ! = 8(7 !) ; also prove that (j + 1 ) ! = (j + 1) (j!).
8. Prove that (j + I ) ! - j! = j(j!).
9.
Verify that 3/2 is a root of 2x4 - 1 3x3 + 27x2 - 4x - 21 = O. Then
denote this polynomial by f(x) and verify Theorem 7.2 by computing the
quotient q(x) in the division off(x) by x - 3/2.
7.2 An Approximation to
a
The underlying reason for the transcendence of a is that it can be
approximated exceptionally well by certain rational numbers. This we
now demonstrate. A good rational approximation to a can be obtained
by taking a finite number of terms from the series (1) that defines a .
We define {3 as the sum of the first j terms of a as given in (1 ) ; that is,
(3)
{3
=
1O- 1 ! + 1O-2 ! + 10- 3 ' + . . . + 1O-j ! .
The value of the integer j will be specified later. We observe that {3 is
rational because it can be written as a sum of fractions, whose denom­
inators are powers of 1 0 ;
\08
N U M BERS : RATIO N A L A N D I R RATION A L
These fractions can all be written with common denominator toi ' , and
so they can be added to give a single fraction,
t
{3 = -' ,
(4)
toi
where the numerator t denotes some integer whose exact value is
immaterial.
The rational number {3 is very close to a. From eqs. (1) and (3), we
see that
a
-
(3 = 1 O-(j+1 ) ! + 10-(;+2)' + 1 O-(;+3 ) ! + . . . .
{3, like that of a itself, consists entirely of
The decimal expansion of a
zeros and ones. The digit 1 appears first in the (j + I) ! place, then in
the (j + 2) ! place, and so on. Thus the number a
{3 is less than
-
-
0.000000 . . . 0000200 ,
where all digits are zeros except the digit 2 in the (j + 1 ) ! place. Another
way of saying this is
(5)
We shall need some other simple inequalities involving a and {3.
Since a and {3 are positive, so are all powers of a and {3. Furthermore,
since a < 1 and {3 < 1 , we see that cI' < 1 , that {¥ < 1, and that
cI'{3-' < 1 for any positive integers r and s, and so we have
(6)
0 < cI' < 1 ,
o < (3s < 1 ,
0 < cI'(¥ < 1 .
7.3 The Plan of the Proof
' To prove that a is transcendental, we shall assume exactly the
reverse, namely that a is algebraic, and we shall obtain a contradiction.
The assumption that a is algebraic means that a satisfies some algebraic
equation with integer coefficients. Among all the algebraic equations
with integer coefficients satisfied by a, select one of lowest degree, say
For brevity we shall write f(x) for the polynomial on the left side of
EX1STENe E 0 F TR A NseEN DENT A L N U M B ER S
109
(7). This polynomial f(x) will play a central role throughout the rest
of the chapter. The basic assumptions about fix) to be kept in mind
are these :
(1) it has integer coefficients ;
(2) the number a is a root ofj(x) = 0, so that f(a) is identically zero
[where by f(a) we mean the result of replacing x by a in f(x)] ;
(3) the number a is a root of no equation of degree less than n, with
integer coefficients.
The number f({3) , obtained by replacing x by (3 in f(x), will also play a
central role in the analysis.
The idea of the proof is this. We will look at the number f(a) - f({3)
[or -f({3) which amounts to the same thing, since f(a) = 0] in two
different ways. One way of looking at -f({3) is as a polynomial in {3,
with integer coefficients. Since (3 is rational, -f({3) is also rational, and
we shall see that its absolute value is relatively large. Another way of
looking at f(a) - f({3) is as the difference of two polynomials, and we
shall show in the next section that this difference has the same order of
magnitude as a - (3, which is relatively small [see eq. (5)]. Thus, by
assuming that a is algebraic, we shall deduce two conflicting orders of
magnitude for f(a) - f({3) and so establish a contradiction.
We shall prepare the way for this in the next section by showing that
( (3) is not zero, and that ( a) - f({3) has the same order of magnitude
as a - /3.
Problem Set 28
I.
Verify the identities :
(a) a4 - {34 = (a - (3) (a3 + a2{3 + a{32 + (33) ;
(b) a5 - {35 = (a - (3) (a4 + a3{3 + a2{32 + a{33 + (34) ;
(c) a6 - {36 = (a - (3) (a5 + a4{3 + a3{32 + a2{33 + a{34 + (35).
2. Write an identity expressing a7 - {37 as a - {3 multiplied by a polynomial
of degree 6.
3. Prove that an algebraic number is a root of infinitely many algebraic
equations with integer coefficients.
7.4 Properties of Polynomials
THEOREM 7 3 . The number (3 is not a root of eq. (7) ; that is, f({3)
.
PROOF. If {3 were a root of (7) then by Theorem 7.2, x
a factor off(x), say
fix) = (x - (3) q(x).
-
-,=
O.
(3 would be
N U M BERS ; RATIONAL A N D I R RATIONAL
1 10
Also by Theorem 7 . 2, q(x) has rational coefficients, and its degree is
one less than the degree of fix). Now, since a is a root of ((x) = 0,
we have
((a)
=
(a - (3) q(a)
=
O.
But this product is zero only if at least one of the factors is zero. The
factor a - {3 is not zero because a is different from (3. Hence q(a) = 0 ;
that is, a is a root of q(x) = 0, and q(x) is of degree n - 1 . If we denote
by k the product of all the denominators of the rational coefficients of
q(x), then the product kq(x) has integer coefficients and a is a root of
kq(x) = O. But this contradicts the fact that a satisfies no equation of
degree less than n, with integer coefficients. Since the assumption that
f({3) = 0 has led to a contradiction, we conclude that f({3) -,= O.
Next, following the outline given in the last section, we show that
/f(a) - f({3)/ is of the same order of magnitude as /a - {3/, which is very
small (see Section 7.2).
THEOREM 7.4. There is a number N, dependent only on the coefficients
off(x) and its degree, such that
If(a)
-
f( {3) / < N (a - (3).
PROOF. The number N is defined by the equation
(8) N = n / c" J + (n
-
1) Je,,- I / + (n
-
2) JC,,-2/ + . . . + 2JC2J + J c d .
Observe, in particular, that N is independent of the integer i used in
the definition of {3.
In the course of the proof, we shall also need the factoring of
ak - 13k and an inequality satisfied by ak - 13k. The factoring is given by
(9)
ak
-
13k
=
(a - (3) ( a'< - I + a'< - 2 {3 + a'<-3 {32 + . . .
where k is any positive integer. This factoring can be verified by multiply­
ing out the right-hand side of (9) :
a{ a'< - I + ak- 2 {3
+ . . . + a{3k-2 + {3k- I )
�
2
= a" + a'<-I {3 + . . . + a2{3f.. - + a - I
EXISTENCE OF TRANSCENDENTAL N U MBERS
111
and
{3(d<-1 + d<-2 {3 + . . . + a{3k-2 + (3k-l )
d<- I {3 + d<-2 {32 + . . . + a{3k- l + {3k .
=
When we subtract these two equations, we observe that all terms, except
the first term of the first equation and the last term of the s�cond equa­
tion, cancel. Thus only ak - {3k remains.
Looking at the right side of eq. (9), we see that each of the terms
ak- 1 , ak-2 {3, etc. is less than 1, by the inequalities (6). B ut since there
are exactly k of these terms and since a - {3 is positive, we can write
( 10)
ak - {3k < (a - (3) (1 + 1 + 1 + . . . + 1 + 1 + 1 )
=
k(a - (3).
Now we compute f(a) and j{(3) using eq. (7) and subtract f({3) from
f(a). Thus we obtain
Next we use the identity (9) to take out the common factor a - (3
from all the terms on the right. This leads to
f( a) - f( {3)
=
(a - (3) [cn(a'I- 1 + a'I-2 {3 + . . . + a{3n- 2 + (3n- l )
. + a{3n-3 + (3n-2)
+ Cn_l(�-2 + �- 3 {3 +
.
.
+ . . . + c d·
Taking absolute values and using Theorem 7.1 and inequality (10), we
get
Observing that l a - {3 1
a
(3 and using the defining equation (8) for
the n umber N, we finally have If(a) - f({3)1 < N(a - (3) and the theorem
is proved.
=
-
7.5 The Transcendence of a
We now complete the proof that the number a defined by eq. (1) is
transcendental. First, we look atf(a) - j{(3) i n another way.
1 12
N U M BERS : RATIO N A L A N D I RRATIONAL
THEOREM 7.5. The number
//(a) - ( 13)/ . 10" i!
( 1 1)
is a positive integer, no matter what value is assigned to the positive
integer j.
PROOF. Since/(a) = 0, the number under discussion can be written as
/ -/(13)/ · 1000 j !
or
//(13)/ · IO" i ' .
From eqs. (7) and (4) we see that
M Ultiplying by l O" j ! we have
/(13) 10" i!
.
and the right-hand side is an integer. This integer cannot be zero, be­
cause J(t3) .,: 0 by Theorem 7.3. Taking absolute values, we see that
or
1/(13)1 . 10" i !
is a positive i nteger, and thus the theorem is proved.
We shall now get an outright contradiction to Theorem 7.5 by show­
ing that the number given by ( 1 1 ) lies between 0 and 1 . In order to do
this, we must choose the integer j, which was used in the definition
of 13, to satisfy
( 1 2)
2N · 1 0" j!
< 1.
10(j+1) !
Can this be done? It can, because this inequality is equivalent to
2N
n j! < 1 ,
(j+1)!
O
1
EXISTENCE OF TRANSCE N DENTAL N U M B ERS
J J3
where the exponent in the denominator can be written as
(j + I ) ! - n . j! = (j + I)j! - n · i ! = (j + I - n) j! .
This exponent can be made as large as we please for fixed n, by taking j
very large. Now n and N are fixed by the equations (7) and (8) ; but
since j depends neither on n nor on N, we can take j so large that ( 1 2)
is satisfied.
Next we show that the number given by ( I I ) lies between 0 and I , by
using Theorem 7.4 and inequality (5), thus :
If( a) - f( 13)1 . 1011 j ' < N(a - 13) · 1 0" j !
2N · IQll j !
< IOU+ 1 ) !
< I,
where i n the last step we used (12). Of course the number i n ( I I ) i s posi­
tive, because of Theorem 7.3.
Hence we have a contradiction, and we conclude that a cannot satisfy
any equation of the form (7). So a is a transcendental number.
7.6 Summary
In this chapter we have answered the q uestion : "Are there any trans­
cendental numbers?" by actually exhibiting a Liouville number and
by proving that it is transcendental, i.e., not algebraic.
Let us recapitulate the entire proof, since the details may have
obscured the argument. We said at the beginning of the chapter that
the central idea is that the number
a
= 1O- 1 ! + 10- 2 ' + 1O-3 ! + 1O-4 ! + . . .
can be approximated very closely by rational numbers. This fact is
stated in the inequality (5) which says, in effect, that a - 13 is very small
compared to 13. We recall that 13 is a rational n umber with denominator
}().I ! [see eq. (4)], but that a- 13 is of the order 1O-(j+ 1 ) ' . In Theorem 7.4,
this small order of magnitude was extended from a - 13 to f( a) - f(l3),
where f(x) is a polynomial with integer coefficients which, for x = a,
allegedly vanishes.
On the other hand, by consideringf( a) - f( 13) in quite a different way
in Theorem 7.5, we showed that the magnitude of f( a) - f( 13) is larger
1 14
N U M B ER S : R ATI O N A L A N D I R R ATI0 N A L
than the earlier estimate. (The factor 1000 j ! in Theorem 7.5 plays no
essential role ; it is present in order to place the two orders of magnitude
of f(ex) - f({3) in sharp contrast.) This was done by recognizing that
f(ex) - f({3) is simply -f({3), andf({3) is a rational number with denomi­
nator 10" j !. Hence the assumption that ex satisfies f(x) = 0 enables us
to prove that f(ex) - f({3) is much larger than the earlier calculation
showed. This contradiction establishes the fact that ex is transcendental.
A P PE N D I X A
Proof That There Are Infinitely
Many Prime Numbers
The argument made here is a so-called indirect proof, otherwise
known as proof by contradiction, or reductio ad absurdum. In this type
of proof, we assume that the proposition is false and then derive a
contradiction from this assumption. Thus, in the case of the present
proposition, we assume that there is only a finite number of primes.
Next we devise a system of notation for the primes. There being only
a finite number, let us denote them by
p . , P2, P3,
. . . , Pk ·
This notation means that there are exactly k primes, where k is some
natural n umber. If we regard these primes as being listed in order of
size, then of course P I = 2, P2 = 3, P3 = 5, P 4 = 7, and so on. Never­
theless, in this proof, it is more convenient to use the notation P h P2,
P3, etc. rather than 2, 3, 5, etc.
Since every natural number can be factored into primes, we observe
that every natural number must be divisible by at least one of the primes
P i t P2 , P3,
.
.
•
, Pk,
because by our assumption there are no primes other than these. But
consider the natural number n which is obtained by multiplying all
the primes and then adding 1 :
n =
P IP2P3
•
•
.
Pk
+
1.
This number n is not divisible by P h because if we divide P I into n we
get a quotient and a remainder with values,
quotient
= P2P3 . . . Pk ,
1 15
remainder = 1 .
t t6
N U M BERS : RATION A L A N D IRRATION A L
If n were divisible by P I > the remainder would be 0 ; thus n is not
divisible by P I .
A similar argument shows that n is not divisible by P2, or P.h or
P 4, . . . , or P/.. ·
We have exhibited a number n which is divisible by no prime what­
ever, and this is an absurd situation. So the assumption that there were
only finitely many primes has led to a logical contradiction, and conse­
quently this assumption must have been false . Hence there are infinitely
many primes.
A P P E N D I X B
Proof of the Fundamental Theorem
of Arithmetic
It is proved in this appendix that every natural number other than 1
can be factored into primes in only one way, except for the order of the
factors. It is understood that any natural number which is itself a
prime, such as 23, is a "factoring into primes" as it stands. Now the
result can be readily checked for small natural numbers. For example,
10 can be factored as 2 . 5, and we know from experience that there is
no other factoring. The same is true of all numbers up to 1 0 :
2
3
4
5
6
7
=
=
=
=
=
=
8 =
9 =
10 =
2
3
2·2
5
2·3
7
2·2·2
3·3
2·5
This list could be continued, but such a listing, however long, could
not be regarded as a proof. For, after all, there are infinitely many
natural numbers, and we cannot check the factoring of them all.
So we must turn to a mathematical argument. The natural numbers
from 2 to 10 have been listed, each with its unique factorization. Now,
either this list can be extended indefinitely so that there is unique
factorization for every natural number, or at some place in the con­
tinued listing the unique factorization property breaks down. These
are the only two possibilities. It is the first of these two possibilities
that we propose to establish, and we shall do it by an indirect argument.
1 17
1 18
N U M B E R S : RATION A L A N D I R RATION A L
We assume that the second possibility holds, i.e., that at some place
in the listing of natural numbers the unique factorization property
breaks down, and show that this leads to a contradiction.
Before carrying out the details of this rather long argument, we give
a brief sketch to guide the reader.
We shall denote the first i nteger which can be factored into primes
in more than one way by m, and we shall write two different prime
factorizations for m. I n Part I of the proof, we shall show that none
of the primes in one factorization of m occurs in the other. Having
established that, if m indeed had two different factorizations, all primes
in one would be different from all primes in the other, we finally con­
struct, in Part I I of the proof, a n umber n which is smaller than m and
also has two different factorizations into primes. This contradicts the
assumption that m was the smallest integer having two different prime
factorizations, and thus completes the proof.
Let us denote by m the first integer which can be factored into
primes in more than one way. In other words we assume that every
natural number smaller than m has the unique factorization property,
whereas m has more than one factoring. Thus we know that there are
at least two different factorizations of m, for which we write
m =
P I P 2P3 . . . Pr
and
What do we mean by this notation ? We mean that m can be factored
i nto primes Ph P2, P3 , and so on, as far as Pro and that there is also
another way of factoring m into the primes q l > q2, Q3, and so on, to qs .
Why not q l > q2, q3 , and so on, to qr ? Because we cannot presume that
the number of prime factors in the two factorings will be the same ; they
might be different for all we know.
The notation needs further explanation. We do not mean, as we did
in Appendix A, that P I is just another label for the prime 2, P2 another
label for the prime 3, and so on. Not at all. We do not know whether
or not the prime 2 is in the batch P I > P2, . . . , Pr' So P I may be 2, or it
may be 23, or it may be 47, or it may be none of these. It is simply
some prime. Similarly, P2 is just some prime. It may be the same prime
as P h or it may not. All that we are presuming is that the natural
number m can be factored into primes in two different ways.
PART I OF THE PROOF. Now, the first thing that we can show is that
the primes P h P 2, . . . , Pr in the first batch are entirely different from
the primes ql> q2, . . . , qs in the second batch. I n other words, if the
prime 7 occurs in the first batch it cannot occur in the second batch.
Since this is not at all obvious, we must give an argument. If the two
A P PE N D I X B
1 19
batches had a prime in common, we could arrange the notation so
that the common prime would be the first one in each batch, thus
P I = q l ' (We may do this because in each factorization the primes can
be thought of in any order.) Now, since P I = q l > we may as well write
PI in place of q l > so that the two factorizations are
m =
P I P2P3 . . . Pr
and
Dividing these equations by P h we obtain
m
- = P2P3
PI
' "
Pr
and
Now we have two different factorings for the natural number m/P I
because we began with two different factorings for m . But this is im­
possible because m was the smallest number having more than one
factoring, and m/P I is smaller than m .
PART II OF THE PROOF. Thus we have established that the primes
in the first factorization of m are entirely different from
the primes qh q2, . . . , qs in the second. In particular, we know that P I
is not equal to q l ; in mathematical symbols, P I .,: ql ' We shall presume
that P I is the smaller of the two ; that is, P I < q l . We have a right to
presume this because the notation is entirely symmetric between the
two batches of primes. Thus, if we can complete the proof in the case
P I < q l > a symmetric proof with the p's and q s interchanged must apply
in an analogous way to the case P I > q l .
Presuming, then, that P I < qh we shall exhibit a number smaller
than m having two different factorings. This will complete the proof
because we shall have contradicted the assumption we made at the
outset, that m was the smallest number with more than one factoring.
A natural number that will meet the stated specifications is
P I > P2, . . . , Pr
'
Note how n is constructed : it is the product of ql
It can be written as a difference,
q2, q3, . . . , q, .
or
- PI
and the primes
1 20
N U M BERS . R A T I O N A L A N D I RRATION A L
and since P l q2Q3 . . . q, is a positive number, this shows that n is smaller
than m .
Finally, we establish that the natural number n has two different
factorings. To do this, we look at the form in which n was introduced,
namely
Each of the factors q2, q3, . . . , q, is a prime, but the first factor, ql - Ph
is not necessarily a prime. If ql - P I were to be factored into primes,
we would have a factoring of n into primes which would not include the
prime P I as one oj the Jactors. To see this, we observe first that the
primes q2, q3, . . , , qs do not have P I among them, as shown in Part I
of the proof. Second, regardless of how ql - P I is factored into primes,
the prime P I could not be present ; for, if P I were a factor in the prime
factorization of ql - P I , then P I would be a divisor o f q l - P I , That is,
the equation
where b is the quotient in the division process, would hold. But this
would lead to the equations
and
ql
=
PI
(l + b),
and the latter can be interpreted as stating that P I is a divisor of q h
which is impossible since no prime can be a divisor of another prime.
Next, we show that n can also be factored in another way so that P I
is one oj the prime factors. To do this, we return to an earlier equation
replace
m
by its form
m =
P I P2P .� . . . P,
and so obtain
n =
P I P2P3 . . . Pr - P l q2q3 . . . qs
=
P I (P 2P 3 . . . Pr - q2q3 ' . . qs) .
The part in parentheses Is not necessarily a prime ; but if we factored it
into primes, we would have a prime factorization of n which includes
APPE N D I X B
121
the prime PI . Thus we have exhibited two factorings o f n or, rather,
two procedures for obtaining factorings of n, one without the prime
P I among the factors and the other with. I n other words, the number
n, whicl-! is smaller than m, has two different prime factorizations. This
completes the proof.
A P P E N D I X C
Cantor's Proof of the Existence of
Transcendental Numbers
I n Chapter 7 we established the existence of transcendental numbers
by exhibiting one. In this appendix, we shall give an independent proof
of their existence by an entirely different method, at the same time
showing that there are infinitely many transcendental numbers. In fact
we establish that in a certain sense there are more transcendental than
algebraic numbers.
At the outset, let us make clear that we are confining our attention
to real algebraic numbers and real transcendental numbers. The roots
of x2 + 1 = 0, for example, are algebraic, but not real algebraic,
numbers. The results we state and their proofs are valid in the complex
case also, but we avoid a few minor complications by restricting our
attention to real numbers.
By a set S we mean any collection of definite, well-distinguished
objects. These objects are called the members of the set S, or the ele­
ments of S. A set S may be finite as, for example, the set of prime
numbers less than 20,
S = p, 3, 5, 7, 1 1 , 1 3, 1 7, 1 9 1 ;
or S may be infinite as, for example, the set of all natural numbers
S
=
1 1 , 2, 3, 4, 5, 6, . . . 1 .
An infinite set is said to be countable (or denumerable) if its members
can be written as a sequence
122
APPENDIX
C
1 23
so that every element of the set is to be found in the sequence. For ex­
ample, the set of even natural numbers can be written as
2, 4, 6, 8, 10, 1 2, . . . ,
so that the nth term in the sequence is 2n, and hence this is a countable
set.
The set of all integers is countable, because it can be written as a
sequence
0,
1, - 1,
2, - 2,
3, - 3,
4, - 4, . . . .
It can be written as a sequence in other ways, but any one way is suffi­
cient to show that the set is countable.
It is not necessary that we know any specific formula for the nth
term of a sequence in order to conclude that a set is countable. For
example, the set of primes
2, 3, 5, 7, 1 1, 1 3 , 1 7, 19, . . .
is countable, even though we do not know the precise value of the
hundred-millionth prime. It is enough to know that there is such a
prime, so that we can conceive of a sequential order for the entire set.
Next, we estab!ish that the set of all rational numbers is countable .
We note that any rational number is a root of a linear equation
ax + b = 0, with integer coefficients a and b. Futhermore we restrict
a to be positive, without any loss of generality. For example, the
rational number 3/5 is a root of 5x - 3 = O. We say that the equation
ax + b = 0 has index
1 + a + Ibl,
so that the index of an equation is a positive integer. For example, the
equation 5x - 3 = 0 has index 9. There is no equation with index I ,
and only one with index 2, namely x = O. Table C 1 includes all linear
equations with indices up to 5. The rational numbers, in order of size,
introduced by the equations of Table C I can also be written in tabular
form, as shown in Table C 2.
It is clear that for any index j there is only a finite number of linear
equations. In fact, there are 2j - 3 equations with index j (the precise
number really has no significance). So with each increasing index only
1 24
N U M BERS : RATIONA L AN 0
I RRATIONAL
TABLE C 1
Index
Equations
2
x=O
3
2x = 0, x + 1 = 0, x - I = °
4
3x = 0, 2x + 1 = 0, 2x - 1 = 0, x + 2 = 0, x - 2 = °
5
4x = 0, 3x + 1 = 0, 3x - 1 = 0, 2x + 2 = 0, 2x - 2 = 0,
x + 3 = 0, x - 3 = °
TABLE C 2
Index
Rational Numbers Introduced
2
°
3
- 1, +1
4
- 2, - t, t, 2
5
- 3, - t, t, 3
a finite number of new rational numbers are introduced. Hence we
can write the rational numbers as a sequence
0, - 1 ,
1
1 , - 2, -"2 '
2'
2, - 3,
:3'
1
-
3'
3, . , . ,
by listing the roots of the equation of index 2, then the roots of all the
equations of index 3, and so on to higher indices, one at a time. Since
every rational number will occur in this sequence, it follows that the
rational numbers are countable.
Virtually the same proof can be used to establish that the set of
algebraic numbers is countable. B ut first we must know something
about how many roots an algebraic equation can have. Recall that an
algebraic number is one which satisfies some equation f(x) = ° of the
type
APPEN D I X C
1 25
with integer coefficients. We may presume that a" is positive, for if it
were negative we could mUltiply the equation by - 1 without affecting
its roots.
THEOREM c. l . Any equation of theform (1) has at most n different roots.
PROOF. Contrary to what is to be proved, let us suppose that eq. (1)
has n + 1 different roots, say f3\. f32 , f33, . . . , f3n, f3n+ l . We now use
Theorem 7.2 (Chapter 7) or, rather, a slight variation on that result.
The proof of that theorem assures us that x - f3 is a factor off(x) if f3
is a root of f(x) = 0, whether or not f3 is a rational number. In case
f3 is irrational, the quotient q(x) has irrational coefficients, but that
does not matter here. Thus in the present context, we see that x - f3 1 is
a factor of f(x), say with quotient ql(X) :
Since (32 is another root of f(x) = 0, we see that it must be a root of
q l (X) = 0, and so x - f32 is a factor of q l (X), say with quotient q2(X) :
q l (X)
f(x)
=
=
(x - f32) qz(x),
(x - (3,) q,(x) = (x - f3,) (x - (:32) q2(X).
Continuing this process with f33 , f34, . . . , (:3", we observe that f(x) can
be factored into
But f(x) is of degree n, so qix) must be a constant ; in fact, qn(x) must
be an in order that this factoring shall agree with eq. ( 1).
Now consider the root f3n+ l , which is different from all the other
roots. From the fact that f(f3n+ l ) = 0, it follows by (2) that
which is impossible since the product of non-zero factors cannot be
zero. Thus Theorem C. 1 is proved.
THEOREM C.2. The set of algebraic numbers is countable.
PROOF. We say that the index of eq. (1) is
N U MBERS :
1 26
RATIONAL A N 0 I R RATION AL
This is a positive i nteger since an is positive, and it is a straightforward
generalization of the definition of index of a linear equation. Again,
we may tabulate all equations for small values of the index, as shown
in Table C 3.
TABLE C 3
I ndex
Equations
2
x = 0
3
4
x2 = 0, 2x = 0, x + 1 = 0, x - I = 0
x3 = 0, 2x2 = 0, x2 + I = 0, x2
I = 0, 3x = 0,
2x + I = 0, 2x
I = 0, x + 2 = 0, x - 2 = 0
-
-
As in the case of linear equations, we now list all new algebraic
numbers arising from the equations of Table C 3. If we take them in
order of magnitude for each index, we obtain the sequence
0 ; - 1 , 1 ; - 2,
2 2, 2 ;
1
-
1
'
- 3,
_
v's
+ 1,
2
-
V2'
_
�
2
(3)
The number 0 comes from the one equation with index 2, the numbers
- 1 and + 1 from the equations with index 3, the numbers - 2, - 1/2,
1 /2, 2 from the equations with index 4, and so on. The number of equa­
tions with any fixed index h is finite, because the degree n and the
coefficients all, . . . , ao are restricted to a finite set of integers. Also, by
Theorem C . I we know that each equation has at most n roots. Hence
the sequence (3) will include all real algebraic numbers. It should be
noted, however, that as we move to higher indices, although we can
at each stage list all equations of any given index, we cannot continue
to list the specific root forms as we have for the first few numbers in (3).
From Theorem C.2 we wish to draw the further conclusion that the
set of real algebraic numbers between 0 and 1 is countable. This follows
from a simple general principle, which we shall formulate as a theorem,
about so-called subsets. A set M is called a subset of a set S if every
element of M is an element of S.
THEOREM C.3. Any infinite subset of a countable set is itself countable.
A PPENDIX C
1 27
PROOF. Let M be an infinite subset of a countable set S, say
S = f a ] , a2 , a3, a4, . 1 . Let ail be the first element of S which is in
M, ai2 the second, and so on. Then M is the set :
.
.
which clearly is countable.
Thus far, every infinite set we have considered has been countable.
We now discuss a contrasting set which is uncountable.
THEOREM C.4. The set of real numbers is uncountable.
PROOF. I n view of Theorem C.3, it will suffice to prove this for real
numbers between 0 and 1 ; specifically for real numbers x satisfying
o < x ;;;i; 1, so that 1 is included and 0 excluded. Suppose that the set
of real numbers between 0 and 1 were countable, say
Write these numbers in decimal form, avoiding terminating decimals
by using the infinite periodic form in all such cases (cf. Section 2.5). For
example, the number 1/2 is to be written as 0.499999 . . , rather than
0.5. Thus we would have
.
We now construct a number
as follows. Let b l be any digit between I and 9, except that b l must be
different from a l l ' Similarly, let b 2 be any non-zero digit other than a22 '
In general, let b" be any non-zero digit other than au . Hence the number
tJ is different from rl (because they differ in the first decimal place),
different from r2 (because they differ in the second decimal place), and,
in general, tJ is different from r" (because they differ in the kth decimal
place). Thus tJ is different from every one of the r's. But {3 is a real
number between 0 and I , and so we have a contradiction.
1 28
N U M BE RS : RAT ION A L A N 0 I R R A T ION A L
From this theorem, we can conclude that since the algebraic numbers
between 0 and 1 are countable, but the real numbers between 0 and 1
are not, there must be real numbers which are not algebraic. These are
the transcendental numbers, whose existence has thus been proved.
THEOREM C.S. The set of real transcendental numbers is uncountable.
PROOF. Suppose the real transcendental numbers were countable, say
Since by Theorem C.2 the real algebraic numbers are countable, say
aJ, a2 , a3, a4, . . , the set of real numbers can be listed sequentially as
.
contrary to Theorem C.4. Thus we have a contradiction and Theorem
C.S is established.
Finally, we note that Theorems C.2 and C.S can be interpreted as
saying that there are "more" transcendental numbers than algebraic
numbers. The algebraic numbers can be listed in an infinite sequence,
but there are too many transcendental numbers to allow such a sequen­
tial listing.
Problem Set 29
1 . (a) List all linear equations with index 6, and (b) list all roots of these
equations that are not roots of linear equations of lower index.
2. Prove that the set of all odd integers, positive and negative, is countable.
3. Prove that the set of polynomials
natural numbers, is countable
a
+ bx4,
where a and b range over all
.
4. List all equations of index 5, and then verify the sequence (3) up to the
element 3.
5. Prove that the set of numbers of the form a + bv3, where a and b range
over all rational numbers, is countable.
6. Prove that, if a set A can be separated into two countable sets B and C,
then A is countable.
7. Prove that the set of real numbers (strictly) between 0 and 0. 1 is uncountable.
8. Prove that the set of all i rrational numbers is uncountable.
Answers and Suggestions
to Selected Problems
Set
1
1 . (a) False : 1 + 1 = 2.
(b) True.
(c) False : l - ( - I) = 2.
(d) True.
* (e) False : 21 + 22 = 6, and 6 is not an integral power of 2.
2. Eight, namely, 1 , 2, 3, 5, 6, 10, 1 5, 30.
3. Five, namely 1, 2, 4, 8, 16.
4. 4.
5. 53, 59, 61, 67, 7 1 , 73, 79, 83, 89, 97.
* 6. Suggestion. Find a good notation for numbers which are exact multiples
of a given number d.
Set
1.
2.
3.
4.
5.
6.
Yes ; q
Yes ; q
Yes ; q
No.
Yes ; q
Yes ; q
= - 7.
= - 7.
= 7.
= - 35.
=
2
7.
8.
9.
10.
11.
No.
Yes ; q = 1 .
No, because q is not unique.
Yes.
Yes.
O.
Set
3
1 . (a), (b), and (f) are true; (c), (d), and (e) are false.
2. True in all cases.
3. (a), (c), and (d) are true; (b) and (e) are false.
4. (a), (b), (c), and (d) are true ; (e) is false.
129
1 30
N U M B E R S : R AT I 0 N A L A N 0 I R R A T I O N AL
Set 4
6. (a) not closed, (b) closed, (c) closed, (d) not closed, (e) closed, (f) closed,
(g) closed.
Set
I . (a) 0.25 ;
(d) 0.01 1 2 ;
(b) 0.01 5 ;
(e) 2.8 1 6 ;
6
(c) 0.8025 ;
(0 1 .2596.
Set
7
Set
8
2. (a) False, for example in case b = 1 0 ;
(b) True ;
(c) False, for example in case b = 1 0 ;
(d) False, for example i n case b = 7 ;
(e) False, for example in case b = 7 ;
( f) True.
3. (a) False, for example in the case of the fraction 3/6;
(b) True ;
(c) False, for example in the case of the fraction 3/6.
4. If ab = 0, then a = 0 or b = O.
5. (b) Yes.
I . (a) 1 /9;
(b) 1 7/3 ;
(c) 3706/9900
= 1 853/4950;
(d) 9978/9990
(e) 1 /9900;
(f) I .
= 1 663/1 665;
Set 9
I . (a) 0. 1 2 ;
(b) 0.3 ;
(c) 4.8 ;
(d) 1 0.0.
2. (a) 0.72999 · . . ; (b) 0.0098999 · . . ; (c) 1 2.999 . . . .
3. Rational numbers alb (in lowest terms) with the property that b is divisible
by no prime other than 2 and 5 and with a � O.
4. None.
Set
7. Rational.
10
A N SWERS A N D S U G G ESTIONS TO PRO B L E M S
1 . v3 and - v'2 will do.
2. V2 and v'2 will do.
Set
11
Set
12
131
v'2 and v'3 will do.
4. v'2 and v'2 will do.
5. v3 and 1/ v'2 will do.
3.
l . (a) n = 3, c3 = 1 5, C2 = - 23, cI = 9, co = - I ;
(b) n = 3, C3 = 3 , C2 = 2, CI = - 3, Co = - 2 ;
(c) n = 3, C3 = 2, C2 = 7, CI = - 3, Co = - 1 8 ;
(d) n = 4, C4 = 2, C 3 = 0, C2 = - I , CI = - 3, Co = 5 ;
(e) n = 5, Cs = 3, C4 = 0, C 3 = - 5, C2 = 6, CI =
1 2 Co = 8 ;
(f) n = 4, C4 = I , C3 = 0, C2 = - 3, C I = - 5, C o = 9.
2. (a) yes ; (b) yes ; (c) yes ; (d) no ; (e) yes ; (f) no.
4. Suggestion. M ultiply the equation by the product b 3b2b l bo.
-
Set
,
13
2 . Suggestion. Use Theorem 4. 1 and one o f the results of Problem 1 .
7. Suggestion. 2/2 i s a root of x2 - 1 = 0, for example.
Set
15
I . (a) Suggestion. Replace () b y 40° i n eq. (5), and use the fact that
cos 1 20° = - 1 /2.
(b) Suggestion. Use the result of Problem I (a) and eq. (8).
(c) Suggestion. Use eq. (8), part I , with () = 10°.
(d) Suggestion. Use the result of Problem I (a) and the identity
cos() = sin (90° - ()) .
3. (a) Suggestion. Replace A by 3() and B by 2() in eq. (I), and use eqs. (3),
(4), (5), and (7).
4. (a), (b), (c), (d), (i), (k) are rational.
Set
16
1 . (a) Suggestion. Use cos 30° = v3/2.
(c) Suggestion. Use cos 45° = v'2/2.
(d) Suggestion. Use the facts that cos 40° is irrational, and that
cos 2 . 35° = cos 70° = cos (90° - 20°) = sin 20°, etc.
3. (b) Yes.
1 32
N U M BERS : RATION A L AN D [ R RATION A L
Set
17
3. Suggestion. Recall that log m + log n
log mn.
4. Suggestion. Make use, among other things, of Example 3 in the text.
=
Set
I . (a) Suggestion. I t i s a root of x2
18
- 3 =
O.
5
O.
(b) Suggestion. It is a root of x3
(c) Suggestion. It is a root of x4 - I Ox2 + 1
(d) Suggestion. See eq. (6) of Chapter 5.
-
Set
=
=
O. See eq.
(5) of Chapter 4.
21
5. (a) False, for example if r = - 2 and s = - 3 ;
(b) False, for example i f r = 4, s = 3 , and c = - 2 ;
(c)
(d)
(e)
(f)
(g)
True ;
True;
True;
False, for example if A = 2/5 ;
True.
6. - 1 0 < A < 10.
8. (b) Yes. The difference is that u -
Set
v
can be 0 in (b) but not in (a).
22
2. (a) 1 , (b) 3, (c) 4, (d) 6, (e) 5, (f) 7, (g) 3, (h) 3 1 , (i) -2, (j) - 22.
Set
1.
2.
3.
• 5.
23
2/1 , 3/2, 5/3, 7/4, 9/5, 10/6, 1 2/7, 1 4/8, 1 6/9, 1 7/10.
3/1 , 6/2, 9/3, 1 3/4, 1 6/5, 1 9/6, 22/7, 25/8, 28/9, 3 1 /10.
Suggestion. Deduce this from Theorem 6.3.
Suggestion. Consider the case where A = v'2 and n = 4, and establish
that there is no fraction m/4 in lowest terms (i.e., with m odd), such that
-!
<
8
A -�
4
Set
1 . n = 4,
2.
n
I
;;;-- 1
m =
!.
8
24
7.
(a) (b) (c) (d) (e)
2
3
<
3
5
4
7
4
7
1
I
(f)
3
4
(g) (h)
5
7
5
7
(i)
5
7
(j) (k) (I) (m) (n)
5
1
1
1
7
7 3 3 3 22
A N S W E R S A N D S U G G E ST I O N S TO P R O B L E M S
Set
1 33
25
I . Suggestion. Take the integer just larger than
A and the integer just smaller.
2. Suggestion. Show that the exception is min with n = 1 , where m is, of the
3.
4.
5.
· 6.
two integers just less than and just greater than A, that one which is
farther away.
(a) 3/2 and 4/3 will do.
(b) 3/2 and 5/3 will do.
(c) 7/3 and 9/4 will do.
(a) All of them.
( b) 1 / 1 , and also 14/10 provided it is taken in the reduced form 7/5.
(a) 3/ 1 , 31 /10, 3 1 4/100; (b) 3/1 .
Suggestion. Prove that the inequalities of Theorem 6.5 are false for
A = 3/5 and any min with n > 5, as follows : A
min is either positive
or negative. If positive, show it is at least 1 /5n ; if negative, at most - 1 /5n.
(a) Suggestion. Use the Fundamental Theorem of Arithmetic, as given in
Appendix B, to prove that the given rational numbers are unequal.
(b) Suggestion. Prove that the inequalities of Theorem 6.5 cannot hold
for any rational number min with n larger than b.
-
· 7.
Set
26
Set
27
1 . (b) There are none.
2. (b) 1 / 1 , 2/1 , 3/2.
3. (b) 1 / 1 , 2/1 .
1.
2.
4.
5.
2, 2, 8, and 10-1.
No.
(b) Ix
71 = x
7 if x � 7; Ix - 71 = -x + 7 if x � 7.
(a) x = - 1 ; (b) x = 2 ; (c) x = 7 and x = - 7 ; (d) all values of x.
-
-
Set
28
2. a7
{37 = (a - (3) (a6 + as{3 + a4{32 + a3{33 + a2{34 + a{3s + (36).
3. Suggestion. Any root off(x) = ° is also a root of f(x) g(x) = 0.
-
Set
29
I . (a) 5 x = 0, 4x ± 1 = 0, 3x ± 2 = 0, 2x ± 3 = 0, x ± 4 = 0 ;
(b) -4, - 3/2, -2/3, - 1 /4, 1 /4, 2/3, 3/2, 4.
.
2. For example, 1 , - 1 , 3, - 3, 5, - 5, 7, -7, 9, -9,
3. Suggestion. Define the index of a + bx4 as a + b ; then observe that there
is only a finite number of polynomials of any given index, and enumerate
them all.
. . ,
1 34
4. x4
N U MBERS : R ATIO N A L A N D I R RATIO N A L
=
2x2
0,
± 1
2x3
=
=
0, x2
0,
±
3
± 1
x
2
=
=
0,
0, 2x2
±
x3
x
±
=
x
=
0, x2
0,
±
±
± 1
x3
x
x2
=
=
0,
0, x2
3x2
± 2x
=
=
0,
0,
3x ± 1
0, 2x ± 2
0, x ± 3
0.
5. Suggestion. All these numbers are algebraic ; use Theorem C.3.
6. Suggestion. Let h I , h2 , h3, . be a sequential listing of the elements of B,
and let c t , C2, C3,
be such a listing of the elements of C: then A can be
written sequentially as
4x
=
0,
=
=
.
=
.
• • •
7. Suggestion. Follow the proof of Theorem C.4; but the numbers a l l , a2 l .
a 3 ) , . . . are all zero in the present context. Construct the desired not-listed
number by choosing
and, in general, hi ¢
hI
=
0, h 2 ¢ al 2
ai - I . i
and
hi ¢ °
and
h 2 ¢ 0, h3 ¢ a2 3
and
h3 ¢ 0,
INOEX
Factor theorem, 106
Fraction, 22
Fundamental theorem of arithmetic,
Absolute value, 1 04
Algebraic number, 71
Approximation, 81 If.
1 2, 1 1 7 If.
Beckenbach, E., 1 04
Bellman, R . , 1 04
Cantor, G., 5, 1 22
Closed under, 9
Commensurable, 45
Complex numbers, 46
Constructions, geometric,
Converse, 29
Countable, 1 22 If.
Courant, Richard, 75
Gelfond, A., 72
Geometric problems,
Hilbert, David,
71
If and only if, meaning of, 26
Incommensurable, 46
Indirect proof, 28
Inequalities, 8 1
Infinite decimals, 23 If., 3 1 , 40
Integer, 1 3
Even, 1 5
Odd, 15
I rrational numbers, 23, 39 If., 52 If.,
73
Decimals, 23
Infinite, 23 If., 3 1 , 40
Non-terminating, 24
Periodic, 31
Terminating, 23
Denumerable, 122 If.
Direct proof, 28
Dirichlet pigeon-hole principle,
Dividend, 21
Division by zero, 1 5
Divisor, 10, 21
Domain, 16
Duplicating the cube, 73
46, 73
65 If.
Irrationality, 42, 52 If., 65 If.
Of logarithmic numbers, 69
Of trigonometric numbers, 651f.
Of v'2, 42
Of v'3, 43
Of v'6, 44
Of v'2 + v'3, 44
93
Liouville, J., 5, 1 04
Logarithmic numbers,
Factor, 10
Factorial, 1 03
Multiple,
1 35
10
65, 69
1 36
I NDEX
Natural numbers, 9
Non-terminating decimals, 24
Number,
Algebraic, 7 1
Complex, 46
Irrational, 39, 52 ff., 69 ff.
Liouville, 104
Logarithmic, 65, 69
Natural, 9
Prime, 1 0
Rational, 2 1
Real, 5, 39
Transcendental, 7 1
Trigonometric, 65
Periodic decimals, 3 1
Pigeon-hole principle, 93
Polynomials, 55, 109
Polynomial equations, 55
Index of, 1 23, 1 25
Primes, 10
Infinite number of, 1 1 5
Proof, the nature of, 1 9, 26 If.
Proportion, 47
Quotient,
Radicals,
10, 2 1
46
Rational fraction, 21
R ational number, 21
Real line, 39
Real numbers, 5, 39
Reciprocal, 23
Remainder, 2 1
Robbins, Herbert, 75
Roots of polynomial equations,
57
Roth, K. E,
104
Schneider, Th., 72
Siegel, C. L., 104
Squaring the circle, 73
Terminating decimals, 23
Thue, A., 104
Transcendence, 71
Of log 2, 7 1
O f 71', 7 1
O f 2v� 7 1
Transcendental numbers, 7 1
Existence of, 1031f., 1 22 If.
Triangle inequality, 105
Trigonometric numbers, 65 If.
Trisection of an angle, 73
Unique factorization,
1 1 , 1 1 7 If.
56,